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Position Vector A position vector r \text{ }\mathbf{r} r of a point P ( x , y , z ) \text{ }P(x, y, z) P ( x , y , z ) is given by: r = x i + y j + z k \text{ }\mathbf{r} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k} r = x i + y j + z k
Example Question Given a point A ( 3 , − 2 , 5 ) \text{ }A(3, -2, 5) A ( 3 , − 2 , 5 ) , find the position vector r \text{ }\mathbf{r} r of the point.
Solution:
The coordinates of point A \text{ }A A are ( 3 , − 2 , 5 ) \text{ }(3, -2, 5) ( 3 , − 2 , 5 ) . Therefore, the position vector r \text{ }\mathbf{r} r is:
r = 3 i − 2 j + 5 k \text{ }\mathbf{r} = 3\mathbf{i} - 2\mathbf{j} + 5\mathbf{k} r = 3 i − 2 j + 5 k
Magnitude of a Position Vector The magnitude of a position vector r \text{ }\mathbf{r} r for a point P ( x , y , z ) \text{ }P(x, y, z) P ( x , y , z ) is calculated as: ∣ r ∣ = x 2 + y 2 + z 2 \text{ }|\mathbf{r}| = \sqrt{x^2 + y^2 + z^2} ∣ r ∣ = x 2 + y 2 + z 2
Example Question Given a point Q ( 2 , − 3 , 4 ) \text{ }Q(2, -3, 4) Q ( 2 , − 3 , 4 ) , find the magnitude of the position vector r \text{ }\mathbf{r} r .
Solution:
For point Q \text{ }Q Q with coordinates ( 2 , − 3 , 4 ) \text{ }(2, -3, 4) ( 2 , − 3 , 4 ) , the magnitude of r \text{ }\mathbf{r} r is:
∣ r ∣ = 2 2 + ( − 3 ) 2 + 4 2 = 4 + 9 + 16 = 29 \text{ }|\mathbf{r}| = \sqrt{2^2 + (-3)^2 + 4^2} = \sqrt{4 + 9 + 16} = \sqrt{29} ∣ r ∣ = 2 2 + ( − 3 ) 2 + 4 2 = 4 + 9 + 16 = 29
Direction Cosines The direction cosines of a vector r \text{ }\mathbf{r} r for a point P ( x , y , z ) \text{ }P(x, y, z) P ( x , y , z ) are given by: cos α = x ∣ r ∣ , cos β = y ∣ r ∣ , cos γ = z ∣ r ∣ \text{ }\cos \alpha = \frac{x}{|\mathbf{r}|}, \quad \cos \beta = \frac{y}{|\mathbf{r}|}, \quad \cos \gamma = \frac{z}{|\mathbf{r}|} cos α = ∣ r ∣ x , cos β = ∣ r ∣ y , cos γ = ∣ r ∣ z They satisfy the relation: cos 2 α + cos 2 β + cos 2 γ = 1 \text{ }\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 cos 2 α + cos 2 β + cos 2 γ = 1
Example Question Given a point R ( 3 , 1 , − 2 ) \text{ }R(3, 1, -2) R ( 3 , 1 , − 2 ) , calculate the direction cosines of the position vector r \text{ }\mathbf{r} r .
Solution:
For point R \text{ }R R with coordinates ( 3 , 1 , − 2 ) \text{ }(3, 1, -2) ( 3 , 1 , − 2 ) , the direction cosines of r \text{ }\mathbf{r} r are:
cos α = 3 ∣ r ∣ = 3 3 2 + 1 2 + ( − 2 ) 2 = 3 14 \text{ }\cos \alpha = \frac{3}{|\mathbf{r}|} = \frac{3}{\sqrt{3^2 + 1^2 + (-2)^2}} = \frac{3}{\sqrt{14}} cos α = ∣ r ∣ 3 = 3 2 + 1 2 + ( − 2 ) 2 3 = 14 3
cos β = 1 ∣ r ∣ = 1 14 \text{ }\cos \beta = \frac{1}{|\mathbf{r}|} = \frac{1}{\sqrt{14}} cos β = ∣ r ∣ 1 = 14 1
cos γ = − 2 ∣ r ∣ = − 2 14 \text{ }\cos \gamma = \frac{-2}{|\mathbf{r}|} = \frac{-2}{\sqrt{14}} cos γ = ∣ r ∣ − 2 = 14 − 2
Addition of Vectors The sum of vectors A \text{ }\mathbf{A} A and B \text{ }\mathbf{B} B is given by: A + B = ( A x + B x ) i + ( A y + B y ) j + ( A z + B z ) k \text{ }\mathbf{A} + \mathbf{B} = (A_x + B_x)\mathbf{i} + (A_y + B_y)\mathbf{j} + (A_z + B_z)\mathbf{k} A + B = ( A x + B x ) i + ( A y + B y ) j + ( A z + B z ) k
Example Question Given vectors A = 2 i − 3 j + 4 k \text{ }\mathbf{A} = 2\mathbf{i} - 3\mathbf{j} + 4\mathbf{k} A = 2 i − 3 j + 4 k and B = − 1 i + 5 j − 2 k \text{ }\mathbf{B} = -1\mathbf{i} + 5\mathbf{j} - 2\mathbf{k} B = − 1 i + 5 j − 2 k , find A + B \text{ }\mathbf{A} + \mathbf{B} A + B .
Solution:
For vectors A \text{ }\mathbf{A} A and B \text{ }\mathbf{B} B :
A + B = ( 2 − 1 ) i + ( − 3 + 5 ) j + ( 4 − 2 ) k = i + 2 j + 2 k \text{ }\mathbf{A} + \mathbf{B} = (2 - 1)\mathbf{i} + (-3 + 5)\mathbf{j} + (4 - 2)\mathbf{k} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k} A + B = ( 2 − 1 ) i + ( − 3 + 5 ) j + ( 4 − 2 ) k = i + 2 j + 2 k
Subtraction of Vectors The difference of vectors A \text{ }\mathbf{A} A and B \text{ }\mathbf{B} B is given by: A − B = ( A x − B x ) i + ( A y − B y ) j + ( A z − B z ) k \text{ }\mathbf{A} - \mathbf{B} = (A_x - B_x)\mathbf{i} + (A_y - B_y)\mathbf{j} + (A_z - B_z)\mathbf{k} A − B = ( A x − B x ) i + ( A y − B y ) j + ( A z − B z ) k
Example Question Given vectors A = 2 i − 3 j + 4 k \text{ }\mathbf{A} = 2\mathbf{i} - 3\mathbf{j} + 4\mathbf{k} A = 2 i − 3 j + 4 k and B = − 1 i + 5 j − 2 k \text{ }\mathbf{B} = -1\mathbf{i} + 5\mathbf{j} - 2\mathbf{k} B = − 1 i + 5 j − 2 k , find A − B \text{ }\mathbf{A} - \mathbf{B} A − B .
Solution:
For vectors A \text{ }\mathbf{A} A and B \text{ }\mathbf{B} B :
A − B = ( 2 + 1 ) i + ( − 3 − 5 ) j + ( 4 + 2 ) k = 3 i − 8 j + 6 k \text{ }\mathbf{A} - \mathbf{B} = (2 + 1)\mathbf{i} + (-3 - 5)\mathbf{j} + (4 + 2)\mathbf{k} = 3\mathbf{i} - 8\mathbf{j} + 6\mathbf{k} A − B = ( 2 + 1 ) i + ( − 3 − 5 ) j + ( 4 + 2 ) k = 3 i − 8 j + 6 k
Multiplication by a Scalar Multiplying vector A \text{ }\mathbf{A} A by a scalar λ \text{ }\lambda λ results in: λ A = λ A x i + λ A y j + λ A z k \text{ }\lambda \mathbf{A} = \lambda A_x \mathbf{i} + \lambda A_y \mathbf{j} + \lambda A_z \mathbf{k} λ A = λ A x i + λ A y j + λ A z k
Example Question Given vector A = 3 i − 2 j + 5 k \text{ }\mathbf{A} = 3\mathbf{i} - 2\mathbf{j} + 5\mathbf{k} A = 3 i − 2 j + 5 k , compute 2 A \text{ }2\mathbf{A} 2 A .
Solution:
For vector A \text{ }\mathbf{A} A :
2 A = 2 ⋅ ( 3 i − 2 j + 5 k ) = 6 i − 4 j + 10 k \text{ }2\mathbf{A} = 2 \cdot (3\mathbf{i} - 2\mathbf{j} + 5\mathbf{k}) = 6\mathbf{i} - 4\mathbf{j} + 10\mathbf{k} 2 A = 2 ⋅ ( 3 i − 2 j + 5 k ) = 6 i − 4 j + 10 k
Unit Vector The unit vector A ^ \text{ }\hat{\mathbf{A}} A ^ of vector A \text{ }\mathbf{A} A is calculated as: A ^ = A ∣ A ∣ \text{ }\hat{\mathbf{A}} = \frac{\mathbf{A}}{|\mathbf{A}|} A ^ = ∣ A ∣ A
Example Question Given vector B = 2 i + 3 j − k \text{ }\mathbf{B} = 2\mathbf{i} + 3\mathbf{j} - \mathbf{k} B = 2 i + 3 j − k , find the unit vector B ^ \text{ }\hat{\mathbf{B}} B ^ .
Solution:
For vector B \text{ }\mathbf{B} B :
∣ B ∣ = 2 2 + 3 2 + ( − 1 ) 2 = 14 \text{ }|\mathbf{B}| = \sqrt{2^2 + 3^2 + (-1)^2} = \sqrt{14} ∣ B ∣ = 2 2 + 3 2 + ( − 1 ) 2 = 14
B ^ = 2 i + 3 j − k 14 \text{ }\hat{\mathbf{B}} = \frac{2\mathbf{i} + 3\mathbf{j} - \mathbf{k}}{\sqrt{14}} B ^ = 14 2 i + 3 j − k
Section Formula If a point P \text{ }P P divides the line segment joining A ( x 1 , y 1 , z 1 ) \text{ }A(x_1, y_1, z_1) A ( x 1 , y 1 , z 1 ) and B ( x 2 , y 2 , z 2 ) \text{ }B(x_2, y_2, z_2) B ( x 2 , y 2 , z 2 ) in the ratio m : n \text{ }m:n m : n , then the coordinates of point P \text{ }P P are given by: P = ( m x 2 + n x 1 m + n , m y 2 + n y 1 m + n , m z 2 + n z 1 m + n ) \text{ }P = \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}, \frac{mz_2 + nz_1}{m+n} \right) P = ( m + n m x 2 + n x 1 , m + n m y 2 + n y 1 , m + n m z 2 + n z 1 )
Example Question Point P \text{ }P P divides the segment joining A ( 1 , 2 , 3 ) \text{ }A(1, 2, 3) A ( 1 , 2 , 3 ) and B ( 4 , − 1 , 5 ) \text{ }B(4, -1, 5) B ( 4 , − 1 , 5 ) in the ratio 2 : 1 \text{ }2:1 2 : 1 . Find the coordinates of point P \text{ }P P .
Solution:
For points A ( 1 , 2 , 3 ) \text{ }A(1, 2, 3) A ( 1 , 2 , 3 ) and B ( 4 , − 1 , 5 ) \text{ }B(4, -1, 5) B ( 4 , − 1 , 5 ) , and ratio 2 : 1 \text{ }2:1 2 : 1 :
P = ( 2 ⋅ 4 + 1 ⋅ 1 2 + 1 , 2 ⋅ ( − 1 ) + 1 ⋅ 2 2 + 1 , 2 ⋅ 5 + 1 ⋅ 3 2 + 1 ) = ( 8 + 1 3 , − 2 + 2 3 , 10 + 3 3 ) = ( 9 3 , 0 , 13 3 ) = ( 3 , 0 , 13 3 ) \text{ }P = \left( \frac{2 \cdot 4 + 1 \cdot 1}{2+1}, \frac{2 \cdot (-1) + 1 \cdot 2}{2+1}, \frac{2 \cdot 5 + 1 \cdot 3}{2+1} \right) = \left( \frac{8 + 1}{3}, \frac{-2 + 2}{3}, \frac{10 + 3}{3} \right) = \left( \frac{9}{3}, 0, \frac{13}{3} \right) = (3, 0, \frac{13}{3}) P = ( 2 + 1 2 ⋅ 4 + 1 ⋅ 1 , 2 + 1 2 ⋅ ( − 1 ) + 1 ⋅ 2 , 2 + 1 2 ⋅ 5 + 1 ⋅ 3 ) = ( 3 8 + 1 , 3 − 2 + 2 , 3 10 + 3 ) = ( 3 9 , 0 , 3 13 ) = ( 3 , 0 , 3 13 )
Midpoint of a Trapezium For trapezium A B C D \text{ }ABCD A BC D with parallel sides A B \text{ }AB A B and C D \text{ }CD C D , the midpoint E \text{ }E E of A B \text{ }AB A B : O E → = O A → + O B → 2 \text{ }\overrightarrow{OE} = \frac{\overrightarrow{OA} + \overrightarrow{OB}}{2} OE = 2 O A + OB The line joining E \text{ }E E and F \text{ }F F is parallel to A B \text{ }AB A B and C D \text{ }CD C D and is half of their sum: E F → = O A → + B C → 2 \text{ }\overrightarrow{EF} = \frac{\overrightarrow{OA} + \overrightarrow{BC}}{2} EF = 2 O A + BC
Example Question Given trapezium A B C D \text{ }ABCD A BC D where O ( 0 , 0 , 0 ) \text{ }O(0, 0, 0) O ( 0 , 0 , 0 ) , A ( 2 , 3 , 1 ) \text{ }A(2, 3, 1) A ( 2 , 3 , 1 ) , B ( 5 , 1 , 4 ) \text{ }B(5, 1, 4) B ( 5 , 1 , 4 ) , and C ( 3 , 4 , 6 ) \text{ }C(3, 4, 6) C ( 3 , 4 , 6 ) , find the coordinates of midpoint E \text{ }E E of side A B \text{ }AB A B and the line joining E \text{ }E E and F \text{ }F F .
Solution:
For points O ( 0 , 0 , 0 ) \text{ }O(0, 0, 0) O ( 0 , 0 , 0 ) , A ( 2 , 3 , 1 ) \text{ }A(2, 3, 1) A ( 2 , 3 , 1 ) , B ( 5 , 1 , 4 ) \text{ }B(5, 1, 4) B ( 5 , 1 , 4 ) :
E → = O A → + O B → 2 = ( 2 , 3 , 1 ) + ( 5 , 1 , 4 ) 2 = ( 7 , 4 , 5 ) 2 = ( 3.5 , 2 , 2.5 ) \text{ }\overrightarrow{E} = \frac{\overrightarrow{OA} + \overrightarrow{OB}}{2} = \frac{(2, 3, 1) + (5, 1, 4)}{2} = \frac{(7, 4, 5)}{2} = (3.5, 2, 2.5) E = 2 O A + OB = 2 ( 2 , 3 , 1 ) + ( 5 , 1 , 4 ) = 2 ( 7 , 4 , 5 ) = ( 3.5 , 2 , 2.5 )
The line joining E \text{ }E E and F \text{ }F F where F ( 3 , 4 , 6 ) \text{ }F(3, 4, 6) F ( 3 , 4 , 6 ) is:
E F → = O A → + B C → 2 = ( 2 , 3 , 1 ) + ( 3 , 4 , 6 ) 2 = ( 5 , 7 , 7 ) 2 = ( 2.5 , 3.5 , 3.5 ) \text{ }\overrightarrow{EF} = \frac{\overrightarrow{OA} + \overrightarrow{BC}}{2} = \frac{(2, 3, 1) + (3, 4, 6)}{2} = \frac{(5, 7, 7)}{2} = (2.5, 3.5, 3.5) EF = 2 O A + BC = 2 ( 2 , 3 , 1 ) + ( 3 , 4 , 6 ) = 2 ( 5 , 7 , 7 ) = ( 2.5 , 3.5 , 3.5 )
Scalar Product (Dot Product) of Vectors: Given vectors A → \text{ }\overrightarrow{A} A and B → \text{ }\overrightarrow{B} B : A → ⋅ B → = A x ⋅ B x + A y ⋅ B y + A z ⋅ B z \text{ }\overrightarrow{A} \cdot \overrightarrow{B} = A_x \cdot B_x + A_y \cdot B_y + A_z \cdot B_z A ⋅ B = A x ⋅ B x + A y ⋅ B y + A z ⋅ B z
Example Question Find the scalar product of vectors A → = ( 2 , − 3 , 5 ) \text{ }\overrightarrow{A} = (2, -3, 5) A = ( 2 , − 3 , 5 ) and B → = ( 4 , 1 , − 2 ) \text{ }\overrightarrow{B} = (4, 1, -2) B = ( 4 , 1 , − 2 ) .
Solution:
Given vectors A → = ( 2 , − 3 , 5 ) \text{ }\overrightarrow{A} = (2, -3, 5) A = ( 2 , − 3 , 5 ) and B → = ( 4 , 1 , − 2 ) \text{ }\overrightarrow{B} = (4, 1, -2) B = ( 4 , 1 , − 2 ) :
A → ⋅ B → = 2 ⋅ 4 + ( − 3 ) ⋅ 1 + 5 ⋅ ( − 2 ) = 8 − 3 − 10 = − 5 \text{ }\overrightarrow{A} \cdot \overrightarrow{B} = 2 \cdot 4 + (-3) \cdot 1 + 5 \cdot (-2) = 8 - 3 - 10 = -5 A ⋅ B = 2 ⋅ 4 + ( − 3 ) ⋅ 1 + 5 ⋅ ( − 2 ) = 8 − 3 − 10 = − 5
Angle Between Two Vectors: Given vectors A → \text{ }\overrightarrow{A} A and B → \text{ }\overrightarrow{B} B : cos θ = A → ⋅ B → ∣ A → ∣ ∣ B → ∣ \text{ }\cos \theta = \frac{\overrightarrow{A} \cdot \overrightarrow{B}}{|\overrightarrow{A}| |\overrightarrow{B}|} cos θ = ∣ A ∣∣ B ∣ A ⋅ B
Example Question Find the angle between vectors A → = ( 3 , − 1 , 2 ) \text{ }\overrightarrow{A} = (3, -1, 2) A = ( 3 , − 1 , 2 ) and B → = ( 2 , 4 , − 3 ) \text{ }\overrightarrow{B} = (2, 4, -3) B = ( 2 , 4 , − 3 ) .
Solution:
Given vectors A → = ( 3 , − 1 , 2 ) \text{ }\overrightarrow{A} = (3, -1, 2) A = ( 3 , − 1 , 2 ) and B → = ( 2 , 4 , − 3 ) \text{ }\overrightarrow{B} = (2, 4, -3) B = ( 2 , 4 , − 3 ) :
∣ A → ∣ = 3 2 + ( − 1 ) 2 + 2 2 = 9 + 1 + 4 = 14 \text{ }|\overrightarrow{A}| = \sqrt{3^2 + (-1)^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14} ∣ A ∣ = 3 2 + ( − 1 ) 2 + 2 2 = 9 + 1 + 4 = 14
∣ B → ∣ = 2 2 + 4 2 + ( − 3 ) 2 = 4 + 16 + 9 = 29 \text{ }|\overrightarrow{B}| = \sqrt{2^2 + 4^2 + (-3)^2} = \sqrt{4 + 16 + 9} = \sqrt{29} ∣ B ∣ = 2 2 + 4 2 + ( − 3 ) 2 = 4 + 16 + 9 = 29
A → ⋅ B → = 3 ⋅ 2 + ( − 1 ) ⋅ 4 + 2 ⋅ ( − 3 ) = 6 − 4 − 6 = − 4 \text{ }\overrightarrow{A} \cdot \overrightarrow{B} = 3 \cdot 2 + (-1) \cdot 4 + 2 \cdot (-3) = 6 - 4 - 6 = -4 A ⋅ B = 3 ⋅ 2 + ( − 1 ) ⋅ 4 + 2 ⋅ ( − 3 ) = 6 − 4 − 6 = − 4
cos θ = − 4 14 ⋅ 29 ≈ − 0.305 \text{ }\cos \theta = \frac{-4}{\sqrt{14} \cdot \sqrt{29}} \approx -0.305 cos θ = 14 ⋅ 29 − 4 ≈ − 0.305
θ = cos − 1 ( − 0.305 ) ≈ 108.8 7 ∘ \text{ }\theta = \cos^{-1}(-0.305) \approx 108.87^\circ θ = cos − 1 ( − 0.305 ) ≈ 108.8 7 ∘
Projection of Vector A → \text{ }\overrightarrow{A} A on Vector B → \text{ }\overrightarrow{B} B : proj B → A → = A → ⋅ B → ∣ B → ∣ 2 B → \text{ }\text{proj}_{\overrightarrow{B}} \overrightarrow{A} = \frac{\overrightarrow{A} \cdot \overrightarrow{B}}{|\overrightarrow{B}|^2} \overrightarrow{B} proj B A = ∣ B ∣ 2 A ⋅ B B
Example Question Find the projection of vector A → = ( 1 , 2 , 3 ) \text{ }\overrightarrow{A} = (1, 2, 3) A = ( 1 , 2 , 3 ) on vector B → = ( 2 , − 1 , 4 ) \text{ }\overrightarrow{B} = (2, -1, 4) B = ( 2 , − 1 , 4 ) .
Solution:
Given vectors A → = ( 1 , 2 , 3 ) \text{ }\overrightarrow{A} = (1, 2, 3) A = ( 1 , 2 , 3 ) and B → = ( 2 , − 1 , 4 ) \text{ }\overrightarrow{B} = (2, -1, 4) B = ( 2 , − 1 , 4 ) :
∣ B → ∣ = 2 2 + ( − 1 ) 2 + 4 2 = 4 + 1 + 16 = 21 \text{ }|\overrightarrow{B}| = \sqrt{2^2 + (-1)^2 + 4^2} = \sqrt{4 + 1 + 16} = \sqrt{21} ∣ B ∣ = 2 2 + ( − 1 ) 2 + 4 2 = 4 + 1 + 16 = 21
A → ⋅ B → = 1 ⋅ 2 + 2 ⋅ ( − 1 ) + 3 ⋅ 4 = 2 − 2 + 12 = 12 \text{ }\overrightarrow{A} \cdot \overrightarrow{B} = 1 \cdot 2 + 2 \cdot (-1) + 3 \cdot 4 = 2 - 2 + 12 = 12 A ⋅ B = 1 ⋅ 2 + 2 ⋅ ( − 1 ) + 3 ⋅ 4 = 2 − 2 + 12 = 12
proj B → A → = 12 21 ⋅ ( 2 , − 1 , 4 ) = ( 24 21 , − 12 21 , 48 21 ) = ( 8 7 , − 4 7 , 16 7 ) \text{ }\text{proj}_{\overrightarrow{B}} \overrightarrow{A} = \frac{12}{21} \cdot (2, -1, 4) = \left( \frac{24}{21}, \frac{-12}{21}, \frac{48}{21} \right) = \left( \frac{8}{7}, -\frac{4}{7}, \frac{16}{7} \right) proj B A = 21 12 ⋅ ( 2 , − 1 , 4 ) = ( 21 24 , 21 − 12 , 21 48 ) = ( 7 8 , − 7 4 , 7 16 )
Vector Product (Cross Product) of Vectors: Given vectors A → \text{ }\overrightarrow{A} A and B → \text{ }\overrightarrow{B} B : A → × B → = ( A y B z − A z B y , A z B x − A x B z , A x B y − A y B x ) \text{ }\overrightarrow{A} \times \overrightarrow{B} = (A_y B_z - A_z B_y, A_z B_x - A_x B_z, A_x B_y - A_y B_x) A × B = ( A y B z − A z B y , A z B x − A x B z , A x B y − A y B x )
Example Question Find the vector product of vectors A → = ( 1 , 2 , − 1 ) \text{ }\overrightarrow{A} = (1, 2, -1) A = ( 1 , 2 , − 1 ) and B → = ( 3 , − 2 , 4 ) \text{ }\overrightarrow{B} = (3, -2, 4) B = ( 3 , − 2 , 4 ) .
Solution:
Given vectors A → = ( 1 , 2 , − 1 ) \text{ }\overrightarrow{A} = (1, 2, -1) A = ( 1 , 2 , − 1 ) and B → = ( 3 , − 2 , 4 ) \text{ }\overrightarrow{B} = (3, -2, 4) B = ( 3 , − 2 , 4 ) :
∣ A → ∣ = 1 2 + 2 2 + ( − 1 ) 2 = 1 + 4 + 1 = 6 \text{ }|\overrightarrow{A}| = \sqrt{1^2 + 2^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6} ∣ A ∣ = 1 2 + 2 2 + ( − 1 ) 2 = 1 + 4 + 1 = 6
∣ B → ∣ = 3 2 + ( − 2 ) 2 + 4 2 = 9 + 4 + 16 = 29 \text{ }|\overrightarrow{B}| = \sqrt{3^2 + (-2)^2 + 4^2} = \sqrt{9 + 4 + 16} = \sqrt{29} ∣ B ∣ = 3 2 + ( − 2 ) 2 + 4 2 = 9 + 4 + 16 = 29
Calculate the components of the cross product:
A → × B → = ( 2 ⋅ 4 − ( − 1 ) ⋅ ( − 2 ) , ( − 1 ) ⋅ 3 − 1 ⋅ 4 , 1 ⋅ ( − 2 ) − 2 ⋅ 3 ) \text{ }\overrightarrow{A} \times \overrightarrow{B} = \left( 2 \cdot 4 - (-1) \cdot (-2), (-1) \cdot 3 - 1 \cdot 4, 1 \cdot (-2) - 2 \cdot 3 \right) A × B = ( 2 ⋅ 4 − ( − 1 ) ⋅ ( − 2 ) , ( − 1 ) ⋅ 3 − 1 ⋅ 4 , 1 ⋅ ( − 2 ) − 2 ⋅ 3 )
A → × B → = ( 8 − 2 , − 3 − 4 , − 2 − 6 ) \text{ }\overrightarrow{A} \times \overrightarrow{B} = (8 - 2, -3 - 4, -2 - 6) A × B = ( 8 − 2 , − 3 − 4 , − 2 − 6 )
A → × B → = ( 6 , − 7 , − 8 ) \text{ }\overrightarrow{A} \times \overrightarrow{B} = (6, -7, -8) A × B = ( 6 , − 7 , − 8 )
Scalar Triple Product: Given vectors A → , B → , C → \text{ }\overrightarrow{A}, \overrightarrow{B}, \overrightarrow{C} A , B , C : A → ⋅ ( B → × C → ) \text{ }\overrightarrow{A} \cdot (\overrightarrow{B} \times \overrightarrow{C}) A ⋅ ( B × C )
Example Question Find the scalar triple product of vectors A → = ( 1 , − 2 , 3 ) \text{ }\overrightarrow{A} = (1, -2, 3) A = ( 1 , − 2 , 3 ) , B → = ( 2 , 1 , − 4 ) \text{ }\overrightarrow{B} = (2, 1, -4) B = ( 2 , 1 , − 4 ) , and C → = ( − 3 , 2 , 5 ) \text{ }\overrightarrow{C} = (-3, 2, 5) C = ( − 3 , 2 , 5 ) .
Solution:
Given vectors A → = ( 1 , − 2 , 3 ) \text{ }\overrightarrow{A} = (1, -2, 3) A = ( 1 , − 2 , 3 ) , B → = ( 2 , 1 , − 4 ) \text{ }\overrightarrow{B} = (2, 1, -4) B = ( 2 , 1 , − 4 ) , and C → = ( − 3 , 2 , 5 ) \text{ }\overrightarrow{C} = (-3, 2, 5) C = ( − 3 , 2 , 5 ) :
B → × C → = ( 1 ⋅ 5 − ( − 4 ) ⋅ 2 , ( − 4 ) ⋅ ( − 3 ) − 1 ⋅ 5 , 1 ⋅ 2 − 2 ⋅ ( − 3 ) ) = ( 5 + 8 , 12 − 5 , 2 + 6 ) = ( 13 , 7 , 8 ) \text{ }\overrightarrow{B} \times \overrightarrow{C} = \left( 1 \cdot 5 - (-4) \cdot 2, (-4) \cdot (-3) - 1 \cdot 5, 1 \cdot 2 - 2 \cdot (-3) \right) = (5 + 8, 12 - 5, 2 + 6) = (13, 7, 8) B × C = ( 1 ⋅ 5 − ( − 4 ) ⋅ 2 , ( − 4 ) ⋅ ( − 3 ) − 1 ⋅ 5 , 1 ⋅ 2 − 2 ⋅ ( − 3 ) ) = ( 5 + 8 , 12 − 5 , 2 + 6 ) = ( 13 , 7 , 8 )
A → ⋅ ( B → × C → ) = 1 ⋅ 13 + ( − 2 ) ⋅ 7 + 3 ⋅ 8 = 13 − 14 + 24 = 23 \text{ }\overrightarrow{A} \cdot (\overrightarrow{B} \times \overrightarrow{C}) = 1 \cdot 13 + (-2) \cdot 7 + 3 \cdot 8 = 13 - 14 + 24 = 23 A ⋅ ( B × C ) = 1 ⋅ 13 + ( − 2 ) ⋅ 7 + 3 ⋅ 8 = 13 − 14 + 24 = 23
Vector Triple Product: Given vectors A → , B → , C → \text{ }\overrightarrow{A}, \overrightarrow{B}, \overrightarrow{C} A , B , C : A → × ( B → × C → ) = ( A → ⋅ C → ) B → − ( A → ⋅ B → ) C → \text{ }\overrightarrow{A} \times (\overrightarrow{B} \times \overrightarrow{C}) = (\overrightarrow{A} \cdot \overrightarrow{C}) \overrightarrow{B} - (\overrightarrow{A} \cdot \overrightarrow{B}) \overrightarrow{C} A × ( B × C ) = ( A ⋅ C ) B − ( A ⋅ B ) C
Example Question Find the vector triple product of vectors A → = ( 1 , 2 , − 1 ) \text{ }\overrightarrow{A} = (1, 2, -1) A = ( 1 , 2 , − 1 ) , B → = ( 3 , − 2 , 4 ) \text{ }\overrightarrow{B} = (3, -2, 4) B = ( 3 , − 2 , 4 ) , and C → = ( 2 , 4 , − 3 ) \text{ }\overrightarrow{C} = (2, 4, -3) C = ( 2 , 4 , − 3 ) .
Solution:
Given vectors A → = ( 1 , 2 , − 1 ) \text{ }\overrightarrow{A} = (1, 2, -1) A = ( 1 , 2 , − 1 ) , B → = ( 3 , − 2 , 4 ) \text{ }\overrightarrow{B} = (3, -2, 4) B = ( 3 , − 2 , 4 ) , and C → = ( 2 , 4 , − 3 ) \text{ }\overrightarrow{C} = (2, 4, -3) C = ( 2 , 4 , − 3 ) :
B → × C → = ( 2 ⋅ 4 − 4 ⋅ ( − 3 ) , ( − 3 ) ⋅ 2 − 4 ⋅ 4 , 1 ⋅ ( − 3 ) − 2 ⋅ 2 ) = ( 8 + 12 , − 6 − 16 , − 3 − 4 ) = ( 20 , − 22 , − 7 ) \text{ }\overrightarrow{B} \times \overrightarrow{C} = \left( 2 \cdot 4 - 4 \cdot (-3), (-3) \cdot 2 - 4 \cdot 4, 1 \cdot (-3) - 2 \cdot 2 \right) = (8 + 12, -6 - 16, -3 - 4) = (20, -22, -7) B × C = ( 2 ⋅ 4 − 4 ⋅ ( − 3 ) , ( − 3 ) ⋅ 2 − 4 ⋅ 4 , 1 ⋅ ( − 3 ) − 2 ⋅ 2 ) = ( 8 + 12 , − 6 − 16 , − 3 − 4 ) = ( 20 , − 22 , − 7 )
A → ⋅ C → = 1 ⋅ 2 + 2 ⋅ 4 + ( − 1 ) ⋅ ( − 3 ) = 2 + 8 + 3 = 13 \text{ }\overrightarrow{A} \cdot \overrightarrow{C} = 1 \cdot 2 + 2 \cdot 4 + (-1) \cdot (-3) = 2 + 8 + 3 = 13 A ⋅ C = 1 ⋅ 2 + 2 ⋅ 4 + ( − 1 ) ⋅ ( − 3 ) = 2 + 8 + 3 = 13
A → ⋅ B → = 1 ⋅ 3 + 2 ⋅ ( − 2 ) + ( − 1 ) ⋅ 4 = 3 − 4 − 4 = − 5 \text{ }\overrightarrow{A} \cdot \overrightarrow{B} = 1 \cdot 3 + 2 \cdot (-2) + (-1) \cdot 4 = 3 - 4 - 4 = -5 A ⋅ B = 1 ⋅ 3 + 2 ⋅ ( − 2 ) + ( − 1 ) ⋅ 4 = 3 − 4 − 4 = − 5
A → × ( B → × C → ) = ( 13 ⋅ ( 3 , − 2 , 4 ) ) − ( − 5 ⋅ ( 2 , 4 , − 3 ) ) = ( 39 , − 26 , 52 ) + ( 10 , − 20 , 15 ) = ( 49 , − 46 , 67 ) \text{ }\overrightarrow{A} \times (\overrightarrow{B} \times \overrightarrow{C}) = (13 \cdot (3, -2, 4)) - (-5 \cdot (2, 4, -3)) = (39, -26, 52) + (10, -20, 15) = (49, -46, 67) A × ( B × C ) = ( 13 ⋅ ( 3 , − 2 , 4 )) − ( − 5 ⋅ ( 2 , 4 , − 3 )) = ( 39 , − 26 , 52 ) + ( 10 , − 20 , 15 ) = ( 49 , − 46 , 67 )
Equation of a Line in Vector Form: A line passing through a point with position vector a → \text{ }\overrightarrow{a} a and parallel to vector b → \text{ }\overrightarrow{b} b : r → = a → + λ b → \text{ }\overrightarrow{r} = \overrightarrow{a} + \lambda \overrightarrow{b} r = a + λ b
Example Question Find the vector equation of the line passing through the point with position vector a → = ( 1 , 2 , 3 ) \text{ }\overrightarrow{a} = (1, 2, 3) a = ( 1 , 2 , 3 ) and parallel to the vector b → = ( 4 , − 1 , 2 ) \text{ }\overrightarrow{b} = (4, -1, 2) b = ( 4 , − 1 , 2 ) .
Solution:
The vector equation of a line passing through a → = ( 1 , 2 , 3 ) \text{ }\overrightarrow{a} = (1, 2, 3) a = ( 1 , 2 , 3 ) and parallel to b → = ( 4 , − 1 , 2 ) \text{ }\overrightarrow{b} = (4, -1, 2) b = ( 4 , − 1 , 2 ) is:
r → = ( 1 , 2 , 3 ) + λ ( 4 , − 1 , 2 ) \text{ }\overrightarrow{r} = (1, 2, 3) + \lambda (4, -1, 2) r = ( 1 , 2 , 3 ) + λ ( 4 , − 1 , 2 )
Equation of a Plane in Vector Form: A plane passing through a point with position vector a → \text{ }\overrightarrow{a} a and normal to vector n → \text{ }\overrightarrow{n} n : r → ⋅ n → = a → ⋅ n → \text{ }\overrightarrow{r} \cdot \overrightarrow{n} = \overrightarrow{a} \cdot \overrightarrow{n} r ⋅ n = a ⋅ n
Example Question Find the vector equation of the plane passing through the point with position vector a → = ( 1 , − 2 , 3 ) \text{ }\overrightarrow{a} = (1, -2, 3) a = ( 1 , − 2 , 3 ) and normal to the vector n → = ( 4 , 5 , − 6 ) \text{ }\overrightarrow{n} = (4, 5, -6) n = ( 4 , 5 , − 6 ) .
Solution:
The vector equation of a plane passing through a → = ( 1 , − 2 , 3 ) \text{ }\overrightarrow{a} = (1, -2, 3) a = ( 1 , − 2 , 3 ) and normal to n → = ( 4 , 5 , − 6 ) \text{ }\overrightarrow{n} = (4, 5, -6) n = ( 4 , 5 , − 6 ) is:
r → ⋅ ( 4 , 5 , − 6 ) = ( 1 , − 2 , 3 ) ⋅ ( 4 , 5 , − 6 ) \text{ }\overrightarrow{r} \cdot (4, 5, -6) = (1, -2, 3) \cdot (4, 5, -6) r ⋅ ( 4 , 5 , − 6 ) = ( 1 , − 2 , 3 ) ⋅ ( 4 , 5 , − 6 )
r → ⋅ ( 4 , 5 , − 6 ) = 4 − 10 − 18 = − 24 \text{ }\overrightarrow{r} \cdot (4, 5, -6) = 4 - 10 - 18 = -24 r ⋅ ( 4 , 5 , − 6 ) = 4 − 10 − 18 = − 24
So the equation of the plane is:
r → ⋅ ( 4 , 5 , − 6 ) = − 24 \text{ }\overrightarrow{r} \cdot (4, 5, -6) = -24 r ⋅ ( 4 , 5 , − 6 ) = − 24
Distance Between Two Points If P ( x 1 , y 1 , z 1 ) \text{ }P(x_1, y_1, z_1) P ( x 1 , y 1 , z 1 ) and Q ( x 2 , y 2 , z 2 ) \text{ }Q(x_2, y_2, z_2) Q ( x 2 , y 2 , z 2 ) are two points in space, the distance P Q \text{ }PQ PQ is given by:
P Q = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 + ( z 2 − z 1 ) 2 \text{ }PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} PQ = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 + ( z 2 − z 1 ) 2
Example Question Find the distance between points P ( 1 , 2 , 3 ) \text{ }P(1, 2, 3) P ( 1 , 2 , 3 ) and Q ( 4 , 6 , 8 ) \text{ }Q(4, 6, 8) Q ( 4 , 6 , 8 ) .
Solution:
The distance P Q \text{ }PQ PQ is given by:
P Q = ( 4 − 1 ) 2 + ( 6 − 2 ) 2 + ( 8 − 3 ) 2 \text{ }PQ = \sqrt{(4 - 1)^2 + (6 - 2)^2 + (8 - 3)^2} PQ = ( 4 − 1 ) 2 + ( 6 − 2 ) 2 + ( 8 − 3 ) 2
P Q = 3 2 + 4 2 + 5 2 \text{ }PQ = \sqrt{3^2 + 4^2 + 5^2} PQ = 3 2 + 4 2 + 5 2
P Q = 9 + 16 + 25 = 50 = 5 2 \text{ }PQ = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2} PQ = 9 + 16 + 25 = 50 = 5 2
So, the distance between the points is 5 2 \text{ }5\sqrt{2} 5 2 units.
Direction Cosines and Ratios If α , β , γ \text{ }\alpha, \beta, \gamma α , β , γ are the angles a vector makes with the positive x \text{ }x x , y \text{ }y y , and z \text{ }z z axes respectively, then the direction cosines are cos α , cos β , cos γ \text{ }\cos\alpha, \cos\beta, \cos\gamma cos α , cos β , cos γ . These satisfy:
cos 2 α + cos 2 β + cos 2 γ = 1 \text{ }\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 cos 2 α + cos 2 β + cos 2 γ = 1
If l , m , n \text{ }l, m, n l , m , n are the direction cosines, then:
l = cos α , m = cos β , n = cos γ \text{ }l = \cos \alpha, \; m = \cos \beta, \; n = \cos \gamma l = cos α , m = cos β , n = cos γ
Example Question Find the direction cosines of a vector making angles 4 5 ∘ , 6 0 ∘ , \text{ }45^\circ, 60^\circ, 4 5 ∘ , 6 0 ∘ , and c o s − 1 ( 1 3 ) \text{ }cos^{-1}\left(\frac{1}{3}\right) co s − 1 ( 3 1 ) with the positive x \text{ }x x , y \text{ }y y , and z \text{ }z z axes respectively.
Solution:
The direction cosines are given by:
l = cos 4 5 ∘ = 1 2 , m = cos 6 0 ∘ = 1 2 , n = cos ( cos − 1 ( 1 3 ) ) = 1 3 \text{ }l = \cos 45^\circ = \frac{1}{\sqrt{2}}, \; m = \cos 60^\circ = \frac{1}{2}, \; n = \cos \left( \cos^{-1} \left(\frac{1}{3}\right) \right) = \frac{1}{3} l = cos 4 5 ∘ = 2 1 , m = cos 6 0 ∘ = 2 1 , n = cos ( cos − 1 ( 3 1 ) ) = 3 1
Thus, the direction cosines are:
l = 1 2 , m = 1 2 , n = 1 3 \text{ }l = \frac{1}{\sqrt{2}}, \; m = \frac{1}{2}, \; n = \frac{1}{3} l = 2 1 , m = 2 1 , n = 3 1
Equation of a Straight Line in Space Vector Form: The equation of a straight line passing through point A ⃗ \vec{A} A with position vector a ⃗ \vec{a} a and parallel to vector b ⃗ \vec{b} b is:
r ⃗ = a ⃗ + t b ⃗ \vec{r} = \vec{a} + t\vec{b} r = a + t b
Cartesian Form: If A ⃗ ( x 1 , y 1 , z 1 ) \vec{A}(x_1, y_1, z_1) A ( x 1 , y 1 , z 1 ) and the line has direction ratios a , b , c a, b, c a , b , c , then the Cartesian equation is:
x − x 1 a = y − y 1 b = z − z 1 c \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} a x − x 1 = b y − y 1 = c z − z 1
Example Questions Example 1: Vector Form Find the vector form equation of a line passing through the point ( 1 , 2 , 3 ) (1, 2, 3) ( 1 , 2 , 3 ) and parallel to vector ( 2 , − 1 , 4 ) (2, -1, 4) ( 2 , − 1 , 4 ) .
Solution:
The vector form equation of a line passing through ( 1 , 2 , 3 ) (1, 2, 3) ( 1 , 2 , 3 ) and parallel to ( 2 , − 1 , 4 ) (2, -1, 4) ( 2 , − 1 , 4 ) is:
r ⃗ = ( 1 , 2 , 3 ) + t ( 2 , − 1 , 4 ) \vec{r} = (1, 2, 3) + t(2, -1, 4) r = ( 1 , 2 , 3 ) + t ( 2 , − 1 , 4 )
Example 2: Cartesian Form Find the Cartesian equation of a line passing through the point ( 1 , 2 , 3 ) (1, 2, 3) ( 1 , 2 , 3 ) and having direction ratios 2 , − 1 , 4 2, -1, 4 2 , − 1 , 4 .
Solution:
The Cartesian equation of a line passing through ( 1 , 2 , 3 ) (1, 2, 3) ( 1 , 2 , 3 ) with direction ratios 2 , − 1 , 4 2, -1, 4 2 , − 1 , 4 is:
x − 1 2 = y − 2 − 1 = z − 3 4 \frac{x - 1}{2} = \frac{y - 2}{-1} = \frac{z - 3}{4} 2 x − 1 = − 1 y − 2 = 4 z − 3
Equation of a Plane Vector Form: The equation of a plane passing through a point A ⃗ ( x 1 , y 1 , z 1 ) \vec{A}(x_1, y_1, z_1) A ( x 1 , y 1 , z 1 ) with normal vector n ⃗ = a i + b j + c k \vec{n} = ai + bj + ck n = ai + bj + c k is:
a ( x − x 1 ) + b ( y − y 1 ) + c ( z − z 1 ) = 0 a(x - x_1) + b(y - y_1) + c(z - z_1) = 0 a ( x − x 1 ) + b ( y − y 1 ) + c ( z − z 1 ) = 0
Cartesian Form: The equation of the plane is:
A x + B y + C z + D = 0 Ax + By + Cz + D = 0 A x + B y + C z + D = 0
Example Question Find the Cartesian equation of the plane passing through the point ( 1 , − 1 , 2 ) (1, -1, 2) ( 1 , − 1 , 2 ) with normal vector ( 2 , 3 , − 4 ) (2, 3, -4) ( 2 , 3 , − 4 ) .
Solution:
The Cartesian equation of the plane passing through ( 1 , − 1 , 2 ) (1, -1, 2) ( 1 , − 1 , 2 ) with normal vector ( 2 , 3 , − 4 ) (2, 3, -4) ( 2 , 3 , − 4 ) is:
2 ( x − 1 ) + 3 ( y + 1 ) − 4 ( z − 2 ) = 0 2(x - 1) + 3(y + 1) - 4(z - 2) = 0 2 ( x − 1 ) + 3 ( y + 1 ) − 4 ( z − 2 ) = 0
2 x − 2 + 3 y + 3 − 4 z + 8 = 0 2x - 2 + 3y + 3 - 4z + 8 = 0 2 x − 2 + 3 y + 3 − 4 z + 8 = 0
2 x + 3 y − 4 z + 9 = 0 2x + 3y - 4z + 9 = 0 2 x + 3 y − 4 z + 9 = 0
Angle Between Two Lines If two lines have direction cosines ( l 1 , m 1 , n 1 ) \text{ }(l_1, m_1, n_1) ( l 1 , m 1 , n 1 ) and ( l 2 , m 2 , n 2 ) \text{ }(l_2, m_2, n_2) ( l 2 , m 2 , n 2 ) , the cosine of the angle θ \text{ }\theta θ between them is given by:
cos θ = l 1 l 2 + m 1 m 2 + n 1 n 2 \text{ }\cos \theta = l_1 l_2 + m_1 m_2 + n_1 n_2 cos θ = l 1 l 2 + m 1 m 2 + n 1 n 2
Example Question Find the cosine of the angle between two lines with direction cosines ( 1 / 3 , 1 / 3 , 1 / 3 ) \text{ }(1/\sqrt{3}, 1/\sqrt{3}, 1/\sqrt{3}) ( 1/ 3 , 1/ 3 , 1/ 3 ) and ( 0 , 1 , 0 ) \text{ }(0, 1, 0) ( 0 , 1 , 0 ) .
Solution:
The cosine of the angle θ \text{ }\theta θ is given by:
cos θ = ( 1 3 ⋅ 0 ) + ( 1 3 ⋅ 1 ) + ( 1 3 ⋅ 0 ) \text{ }\cos \theta = \left( \frac{1}{\sqrt{3}} \cdot 0 \right) + \left( \frac{1}{\sqrt{3}} \cdot 1 \right) + \left( \frac{1}{\sqrt{3}} \cdot 0 \right) cos θ = ( 3 1 ⋅ 0 ) + ( 3 1 ⋅ 1 ) + ( 3 1 ⋅ 0 )
cos θ = 0 + 1 3 + 0 = 1 3 \text{ }\cos \theta = 0 + \frac{1}{\sqrt{3}} + 0 = \frac{1}{\sqrt{3}} cos θ = 0 + 3 1 + 0 = 3 1
Angle Between Two Planes If the normal vectors of two planes are n 1 ⃗ = a 1 i + b 1 j + c 1 k \text{ }\vec{n_1} = a_1i + b_1j + c_1k n 1 = a 1 i + b 1 j + c 1 k and n 2 ⃗ = a 2 i + b 2 j + c 2 k \text{ }\vec{n_2} = a_2i + b_2j + c_2k n 2 = a 2 i + b 2 j + c 2 k , the cosine of the angle θ \text{ }\theta θ between the planes is given by:
cos θ = a 1 a 2 + b 1 b 2 + c 1 c 2 a 1 2 + b 1 2 + c 1 2 ⋅ a 2 2 + b 2 2 + c 2 2 \text{ }\cos \theta = \frac{a_1a_2 + b_1b_2 + c_1c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \cdot \sqrt{a_2^2 + b_2^2 + c_2^2}} cos θ = a 1 2 + b 1 2 + c 1 2 ⋅ a 2 2 + b 2 2 + c 2 2 a 1 a 2 + b 1 b 2 + c 1 c 2
Example Question Find the cosine of the angle between two planes with normal vectors ( 1 , 2 , 3 ) \text{ }(1, 2, 3) ( 1 , 2 , 3 ) and ( 4 , − 5 , 6 ) \text{ }(4, -5, 6) ( 4 , − 5 , 6 ) .
Solution:
The cosine of the angle θ \text{ }\theta θ is given by:
cos θ = 1 ⋅ 4 + 2 ⋅ ( − 5 ) + 3 ⋅ 6 1 2 + 2 2 + 3 2 ⋅ 4 2 + ( − 5 ) 2 + 6 2 \text{ }\cos \theta = \frac{1 \cdot 4 + 2 \cdot (-5) + 3 \cdot 6}{\sqrt{1^2 + 2^2 + 3^2} \cdot \sqrt{4^2 + (-5)^2 + 6^2}} cos θ = 1 2 + 2 2 + 3 2 ⋅ 4 2 + ( − 5 ) 2 + 6 2 1 ⋅ 4 + 2 ⋅ ( − 5 ) + 3 ⋅ 6
cos θ = 4 − 10 + 18 1 + 4 + 9 ⋅ 16 + 25 + 36 \text{ }\cos \theta = \frac{4 - 10 + 18}{\sqrt{1 + 4 + 9} \cdot \sqrt{16 + 25 + 36}} cos θ = 1 + 4 + 9 ⋅ 16 + 25 + 36 4 − 10 + 18
cos θ = 12 14 ⋅ 77 = 12 1078 = 12 32.82 ≈ 0.365 \text{ }\cos \theta = \frac{12}{\sqrt{14} \cdot \sqrt{77}} = \frac{12}{\sqrt{1078}} = \frac{12}{32.82} \approx 0.365 cos θ = 14 ⋅ 77 12 = 1078 12 = 32.82 12 ≈ 0.365
Distance of a Point from a Plane If a point P ( x 1 , y 1 , z 1 ) \text{ }P(x_1, y_1, z_1) P ( x 1 , y 1 , z 1 ) and a plane A x + B y + C z + D = 0 \text{ }Ax + By + Cz + D = 0 A x + B y + C z + D = 0 , the distance d \text{ }d d is given by:
d = ∣ A x 1 + B y 1 + C z 1 + D ∣ A 2 + B 2 + C 2 \text{ }d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} d = A 2 + B 2 + C 2 ∣ A x 1 + B y 1 + C z 1 + D ∣
Example Question Find the distance from the point ( 3 , − 1 , 2 ) \text{ }(3, -1, 2) ( 3 , − 1 , 2 ) to the plane 2 x − 3 y + 6 z + 5 = 0 \text{ }2x - 3y + 6z + 5 = 0 2 x − 3 y + 6 z + 5 = 0 .
Solution:
The distance d \text{ }d d is given by:
d = ∣ 2 ⋅ 3 − 3 ⋅ ( − 1 ) + 6 ⋅ 2 + 5 ∣ 2 2 + ( − 3 ) 2 + 6 2 \text{ }d = \frac{|2 \cdot 3 - 3 \cdot (-1) + 6 \cdot 2 + 5|}{\sqrt{2^2 + (-3)^2 + 6^2}} d = 2 2 + ( − 3 ) 2 + 6 2 ∣2 ⋅ 3 − 3 ⋅ ( − 1 ) + 6 ⋅ 2 + 5∣
d = ∣ 6 + 3 + 12 + 5 ∣ 4 + 9 + 36 \text{ }d = \frac{|6 + 3 + 12 + 5|}{\sqrt{4 + 9 + 36}} d = 4 + 9 + 36 ∣6 + 3 + 12 + 5∣
d = 26 49 = 26 7 ≈ 3.71 \text{ }d = \frac{26}{\sqrt{49}} = \frac{26}{7} \approx 3.71 d = 49 26 = 7 26 ≈ 3.71
So, the distance from the point to the plane is approximately 3.71 \text{ }3.71 3.71 units.
Distance Between Two Parallel Planes If the equations of two parallel planes are A x + B y + C z + D 1 = 0 \text{ }Ax + By + Cz + D_1 = 0 A x + B y + C z + D 1 = 0 and A x + B y + C z + D 2 = 0 \text{ }Ax + By + Cz + D_2 = 0 A x + B y + C z + D 2 = 0 , the distance d \text{ }d d between them is:
d = ∣ D 2 − D 1 ∣ A 2 + B 2 + C 2 \text{ }d = \frac{|D_2 - D_1|}{\sqrt{A^2 + B^2 + C^2}} d = A 2 + B 2 + C 2 ∣ D 2 − D 1 ∣
Example Question Find the distance between the planes 2 x − 3 y + 4 z − 5 = 0 \text{ }2x - 3y + 4z - 5 = 0 2 x − 3 y + 4 z − 5 = 0 and 2 x − 3 y + 4 z + 7 = 0 \text{ }2x - 3y + 4z + 7 = 0 2 x − 3 y + 4 z + 7 = 0 .
Solution:
The distance d \text{ }d d is given by:
d = ∣ 7 − ( − 5 ) ∣ 2 2 + ( − 3 ) 2 + 4 2 \text{ }d = \frac{|7 - (-5)|}{\sqrt{2^2 + (-3)^2 + 4^2}} d = 2 2 + ( − 3 ) 2 + 4 2 ∣7 − ( − 5 ) ∣
d = 12 29 ≈ 12 5.39 ≈ 2.23 \text{ }d = \frac{12}{\sqrt{29}} \approx \frac{12}{5.39} \approx 2.23 d = 29 12 ≈ 5.39 12 ≈ 2.23
So, the distance between the planes is approximately 2.23 \text{ }2.23 2.23 units.
Length of a Vector: Given vector A → \text{ }\overrightarrow{A} A : ∣ A → ∣ = A → ⋅ A → \text{ }|\overrightarrow{A}| = \sqrt{\overrightarrow{A} \cdot \overrightarrow{A}} ∣ A ∣ = A ⋅ A
Example Question Find the length of the vector V ⃗ = ( 3 , − 4 , 5 ) \text{ }\vec{V} = (3, -4, 5) V = ( 3 , − 4 , 5 ) .
Solution:
The length of the vector V ⃗ \text{ }\vec{V} V is:
∣ V ⃗ ∣ = ( 3 ) 2 + ( − 4 ) 2 + ( 5 ) 2 \text{ }|\vec{V}| = \sqrt{(3)^2 + (-4)^2 + (5)^2} ∣ V ∣ = ( 3 ) 2 + ( − 4 ) 2 + ( 5 ) 2
∣ V ⃗ ∣ = 9 + 16 + 25 \text{ }|\vec{V}| = \sqrt{9 + 16 + 25} ∣ V ∣ = 9 + 16 + 25
∣ V ⃗ ∣ = 50 = 5 2 \text{ }|\vec{V}| = \sqrt{50} = 5\sqrt{2} ∣ V ∣ = 50 = 5 2
Properties of Scalar Product: 1. A → ⋅ B → = B → ⋅ A → \text{ }\overrightarrow{A} \cdot \overrightarrow{B} = \overrightarrow{B} \cdot \overrightarrow{A} A ⋅ B = B ⋅ A 2. A → ⋅ ( λ B → ) = λ ( A → ⋅ B → ) \text{ }\overrightarrow{A} \cdot (\lambda \overrightarrow{B}) = \lambda (\overrightarrow{A} \cdot \overrightarrow{B}) A ⋅ ( λ B ) = λ ( A ⋅ B ) 3. A → ⋅ ( B → + C → ) = A → ⋅ B → + A → ⋅ C → \text{ }\overrightarrow{A} \cdot (\overrightarrow{B} + \overrightarrow{C}) = \overrightarrow{A} \cdot \overrightarrow{B} + \overrightarrow{A} \cdot \overrightarrow{C} A ⋅ ( B + C ) = A ⋅ B + A ⋅ C
Example Question Verify the properties of scalar product for vectors A ⃗ = ( 2 , 3 , − 1 ) \text{ }\vec{A} = (2, 3, -1) A = ( 2 , 3 , − 1 ) and B ⃗ = ( 4 , − 2 , 5 ) \text{ }\vec{B} = (4, -2, 5) B = ( 4 , − 2 , 5 ) .
Solution:
Let's verify each property:
A ⃗ ⋅ B ⃗ = ( 2 ) ( 4 ) + ( 3 ) ( − 2 ) + ( − 1 ) ( 5 ) = 8 − 6 − 5 = − 3 \text{ }\vec{A} \cdot \vec{B} = (2)(4) + (3)(-2) + (-1)(5) = 8 - 6 - 5 = -3 A ⋅ B = ( 2 ) ( 4 ) + ( 3 ) ( − 2 ) + ( − 1 ) ( 5 ) = 8 − 6 − 5 = − 3 A ⃗ ⋅ ( 3 B ⃗ ) = ( 2 ) ( 3 ⋅ 4 ) + ( 3 ) ( 3 ⋅ − 2 ) + ( − 1 ) ( 3 ⋅ 5 ) = 24 − 18 − 15 = − 9 \text{ }\vec{A} \cdot (3\vec{B}) = (2)(3 \cdot 4) + (3)(3 \cdot -2) + (-1)(3 \cdot 5) = 24 - 18 - 15 = -9 A ⋅ ( 3 B ) = ( 2 ) ( 3 ⋅ 4 ) + ( 3 ) ( 3 ⋅ − 2 ) + ( − 1 ) ( 3 ⋅ 5 ) = 24 − 18 − 15 = − 9 A ⃗ ⋅ ( B ⃗ + C ⃗ ) = A ⃗ ⋅ B ⃗ + v e c A ⋅ C ⃗ \text{ }\vec{A} \cdot (\vec{B} + \vec{C}) = \vec{A} \cdot \vec{B} + \ vec{A} \cdot \vec{C} A ⋅ ( B + C ) = A ⋅ B + v ec A ⋅ C Let's take another vector C ⃗ = ( − 1 , 2 , 3 ) \text{ }\vec{C} = (-1, 2, 3) C = ( − 1 , 2 , 3 ) . A ⃗ ⋅ ( B ⃗ + C ⃗ ) = ( 2 ) ( 4 − 1 ) + ( 3 ) ( − 2 + 2 ) + ( − 1 ) ( 5 + 3 ) = 6 − 6 − 8 = − 8 \text{ }\vec{A} \cdot (\vec{B} + \vec{C}) = (2)(4 - 1) + (3)(-2 + 2) + (-1)(5 + 3) = 6 - 6 - 8 = -8 A ⋅ ( B + C ) = ( 2 ) ( 4 − 1 ) + ( 3 ) ( − 2 + 2 ) + ( − 1 ) ( 5 + 3 ) = 6 − 6 − 8 = − 8 A ⃗ ⋅ B ⃗ + A ⃗ ⋅ C ⃗ = − 3 + ( 2 ) ( − 1 ) + ( 3 ) ( 2 ) + ( − 1 ) ( 3 ) = − 3 − 2 + 6 − 3 = − 2 \text{ }\vec{A} \cdot \vec{B} + \vec{A} \cdot \vec{C} = -3 + (2)(-1) + (3)(2) + (-1)(3) = -3 - 2 + 6 - 3 = -2 A ⋅ B + A ⋅ C = − 3 + ( 2 ) ( − 1 ) + ( 3 ) ( 2 ) + ( − 1 ) ( 3 ) = − 3 − 2 + 6 − 3 = − 2 Thus, all properties hold true for the given vectors.
Properties of Vector Product: 1. A → × B → = − ( B → × A → ) \text{ }\overrightarrow{A} \times \overrightarrow{B} = - (\overrightarrow{B} \times \overrightarrow{A}) A × B = − ( B × A ) 2. A → × ( B → + C → ) = A → × B → + A → × C → \text{ }\overrightarrow{A} \times (\overrightarrow{B} + \overrightarrow{C}) = \overrightarrow{A} \times \overrightarrow{B} + \overrightarrow{A} \times \overrightarrow{C} A × ( B + C ) = A × B + A × C 3. A → × ( λ B → ) = λ ( A → × B → ) \text{ }\overrightarrow{A} \times (\lambda \overrightarrow{B}) = \lambda (\overrightarrow{A} \times \overrightarrow{B}) A × ( λ B ) = λ ( A × B )
Example Question Verify the properties of vector product for vectors A ⃗ = ( 2 , 1 , − 3 ) \text{ }\vec{A} = (2, 1, -3) A = ( 2 , 1 , − 3 ) and B ⃗ = ( 4 , − 2 , 5 ) \text{ }\vec{B} = (4, -2, 5) B = ( 4 , − 2 , 5 ) .
Solution:
Let's verify each property:
A ⃗ × B ⃗ = ( 2 i + j − 3 k ) × ( 4 i − 2 j + 5 k ) \text{ }\vec{A} \times \vec{B} = (2i + j - 3k) \times (4i - 2j + 5k) A × B = ( 2 i + j − 3 k ) × ( 4 i − 2 j + 5 k ) = ( 1 ) ( 5 ) − ( − 3 ) ( − 2 ) − ( 2 ) ( 4 ) − ( ( − 3 ) ( 4 ) − 2 ( 5 ) − ( 1 ) ( − 2 ) ) i + ( ( 2 ) ( 4 ) − ( 2 ) ( 1 ) − ( 1 ) ( 5 ) ) j + ( ( − 3 ) ( − 2 ) − 1 ( 4 ) − ( 1 ) ( − 2 ) ) k \text{ }= (1)(5) - (-3)(-2) - (2)(4) - ((-3)(4) - 2(5) - (1)(-2))i + ((2)(4) - (2)(1) - (1)(5))j + ((-3)(-2) - 1(4) - (1)(-2))k = ( 1 ) ( 5 ) − ( − 3 ) ( − 2 ) − ( 2 ) ( 4 ) − (( − 3 ) ( 4 ) − 2 ( 5 ) − ( 1 ) ( − 2 )) i + (( 2 ) ( 4 ) − ( 2 ) ( 1 ) − ( 1 ) ( 5 )) j + (( − 3 ) ( − 2 ) − 1 ( 4 ) − ( 1 ) ( − 2 )) k
= 5 + 6 − 8 − ( 8 − 10 − 2 ) i + ( 8 − 2 − 5 ) j + ( 6 − 4 + 2 ) k \text{ }= 5 + 6 - 8 - (8 - 10 - 2)i + (8 - 2 - 5)j + (6 - 4 + 2)k = 5 + 6 − 8 − ( 8 − 10 − 2 ) i + ( 8 − 2 − 5 ) j + ( 6 − 4 + 2 ) k
= 3 i − 1 j + 4 k \text{ }= 3i - 1j + 4k = 3 i − 1 j + 4 k
Now, let's calculate B ⃗ × A ⃗ \text{ }\vec{B} \times \vec{A} B × A .
B ⃗ × A ⃗ = − ( A ⃗ × B ⃗ ) \text{ }\vec{B} \times \vec{A} = - (\vec{A} \times \vec{B}) B × A = − ( A × B )
= − ( 3 i − 1 j + 4 k ) \text{ }= - (3i - 1j + 4k) = − ( 3 i − 1 j + 4 k )
= − 3 i + j − 4 k \text{ }= -3i + j - 4k = − 3 i + j − 4 k
Thus, A ⃗ × B ⃗ = − ( B ⃗ × A ⃗ ) \text{ }\vec{A} \times \vec{B} = - (\vec{B} \times \vec{A}) A × B = − ( B × A ) .
Similarly, let's verify the other properties. A ⃗ × ( B ⃗ + C ⃗ ) = A ⃗ × B ⃗ + A ⃗ × C ⃗ \text{ }\vec{A} \times (\vec{B} + \vec{C}) = \vec{A} \times \vec{B} + \vec{A} \times \vec{C} A × ( B + C ) = A × B + A × C We'll use another vector C ⃗ = ( − 1 , 2 , 3 ) \text{ }\vec{C} = (-1, 2, 3) C = ( − 1 , 2 , 3 ) .
A ⃗ × ( B ⃗ + C ⃗ ) = ( 3 i − 1 j + 4 k ) × ( ( 4 i − 2 j + 5 k ) + ( − i + 2 j + 3 k ) ) \text{ }\vec{A} \times (\vec{B} + \vec{C}) = (3i - 1j + 4k) \times ((4i - 2j + 5k) + (-i + 2j + 3k)) A × ( B + C ) = ( 3 i − 1 j + 4 k ) × (( 4 i − 2 j + 5 k ) + ( − i + 2 j + 3 k ))
= ( 3 i − 1 j + 4 k ) × ( 3 i − j + 8 k ) \text{ }= (3i - 1j + 4k) \times (3i - j + 8k) = ( 3 i − 1 j + 4 k ) × ( 3 i − j + 8 k )
= ( 1 ) ( 8 ) − ( 4 ) ( − 1 ) − ( 3 ) ( 3 ) − ( ( 3 ) ( 3 ) − 4 ( − 1 ) − ( 1 ) ( 8 ) ) i + ( ( 3 ) ( 3 ) − 1 ( 8 ) − 4 ( 8 ) ) j + ( ( 3 ) ( − 1 ) − 1 ( 3 ) − 4 ( 3 ) ) k \text{ }= (1)(8) - (4)(-1) - (3)(3) - ((3)(3) - 4(-1) - (1)(8))i + ((3)(3) - 1(8) - 4(8))j + ((3)(-1) - 1(3) - 4(3))k = ( 1 ) ( 8 ) − ( 4 ) ( − 1 ) − ( 3 ) ( 3 ) − (( 3 ) ( 3 ) − 4 ( − 1 ) − ( 1 ) ( 8 )) i + (( 3 ) ( 3 ) − 1 ( 8 ) − 4 ( 8 )) j + (( 3 ) ( − 1 ) − 1 ( 3 ) − 4 ( 3 )) k
= 8 + 4 − 9 − ( 9 + 4 − 8 ) i + ( 9 − 8 − 32 ) j + ( − 3 − 3 − 12 ) k \text{ }= 8 + 4 - 9 - (9 + 4 - 8)i + (9 - 8 - 32)j + (-3 - 3 - 12)k = 8 + 4 − 9 − ( 9 + 4 − 8 ) i + ( 9 − 8 − 32 ) j + ( − 3 − 3 − 12 ) k
= 3 i − 27 j − 18 k \text{ }= 3i - 27j - 18k = 3 i − 27 j − 18 k
Now, let's calculate A ⃗ × B ⃗ + A ⃗ × C ⃗ \text{ }\vec{A} \times \vec{B} + \vec{A} \times \vec{C} A × B + A × C .
A ⃗ × B ⃗ + A ⃗ × C ⃗ = ( 3 i − 1 j + 4 k ) + ( 3 i − 27 j − 18 k ) \text{ }\vec{A} \times \vec{B} + \vec{A} \times \vec{C} = (3i - 1j + 4k) + (3i - 27j - 18k) A × B + A × C = ( 3 i − 1 j + 4 k ) + ( 3 i − 27 j − 18 k )
= 3 i − 1 j + 4 k + 3 i − 27 j − 18 k \text{ }= 3i - 1j + 4k + 3i - 27j - 18k = 3 i − 1 j + 4 k + 3 i − 27 j − 18 k
= 6 i − 28 j − 14 k \text{ }= 6i - 28j - 14k = 6 i − 28 j − 14 k
Thus, A ⃗ × ( B ⃗ + C ⃗ ) = A ⃗ × B ⃗ + A ⃗ × C ⃗ \text{ }\vec{A} \times (\vec{B} + \vec{C}) = \vec{A} \times \vec{B} + \vec{A} \times \vec{C} A × ( B + C ) = A × B + A × C .
Similarly, let's verify the last property. ... Area of a Triangle: Given vectors A → \text{ }\overrightarrow{A} A and B → \text{ }\overrightarrow{B} B representing two sides of the triangle: Area = 1 2 ∣ A → × B → ∣ \text{ }\text{Area} = \frac{1}{2} |\overrightarrow{A} \times \overrightarrow{B}| Area = 2 1 ∣ A × B ∣
Example Question Find the area of the triangle formed by the vectors A ⃗ = ( 3 , − 4 , 5 ) \text{ }\vec{A} = (3, -4, 5) A = ( 3 , − 4 , 5 ) and B ⃗ = ( 1 , 2 , − 2 ) \text{ }\vec{B} = (1, 2, -2) B = ( 1 , 2 , − 2 ) .
Solution:
The area of the triangle is:
Area = 1 2 ∣ A ⃗ × B ⃗ ∣ \text{ }\text{Area} = \frac{1}{2} |\vec{A} \times \vec{B}| Area = 2 1 ∣ A × B ∣
Let's calculate A ⃗ × B ⃗ \text{ }\vec{A} \times \vec{B} A × B .
A ⃗ × B ⃗ = ( 3 i − 4 j + 5 k ) × ( i + 2 j − 2 k ) \text{ }\vec{A} \times \vec{B} = (3i - 4j + 5k) \times (i + 2j - 2k) A × B = ( 3 i − 4 j + 5 k ) × ( i + 2 j − 2 k )
= ( ( − 4 ) ( − 2 ) − ( 5 ) ( 2 ) ) i + ( ( 5 ) ( 1 ) − ( 3 ) ( − 2 ) ) j + ( ( 3 ) ( 2 ) − ( − 4 ) ( 1 ) ) k \text{ }= ((-4)(-2) - (5)(2))i + ((5)(1) - (3)(-2))j + ((3)(2) - (-4)(1))k = (( − 4 ) ( − 2 ) − ( 5 ) ( 2 )) i + (( 5 ) ( 1 ) − ( 3 ) ( − 2 )) j + (( 3 ) ( 2 ) − ( − 4 ) ( 1 )) k
= ( 8 − 10 ) i + ( 5 + 6 ) j + ( 6 + 4 ) k \text{ }= (8 - 10)i + (5 + 6)j + (6 + 4)k = ( 8 − 10 ) i + ( 5 + 6 ) j + ( 6 + 4 ) k
= − 2 i + 11 j + 10 k \text{ }= -2i + 11j + 10k = − 2 i + 11 j + 10 k
Now, let's find the magnitude of A ⃗ × B ⃗ \text{ }\vec{A} \times \vec{B} A × B .
∣ A ⃗ × B ⃗ ∣ = ( − 2 ) 2 + 1 1 2 + 1 0 2 \text{ }|\vec{A} \times \vec{B}| = \sqrt{(-2)^2 + 11^2 + 10^2} ∣ A × B ∣ = ( − 2 ) 2 + 1 1 2 + 1 0 2
= 4 + 121 + 100 \text{ }= \sqrt{4 + 121 + 100} = 4 + 121 + 100
= 225 = 15 \text{ }= \sqrt{225} = 15 = 225 = 15
Therefore, the area of the triangle is:
Area = 1 2 × 15 = 7.5 \text{ }\text{Area} = \frac{1}{2} \times 15 = 7.5 Area = 2 1 × 15 = 7.5 square units
Volume of a Parallelepiped: Given vectors A → , B → , C → \overrightarrow{A}, \overrightarrow{B}, \overrightarrow{C} A , B , C :Volume = ∣ A → ⋅ ( B → × C → ) ∣ \text{Volume} = |\overrightarrow{A} \cdot (\overrightarrow{B} \times \overrightarrow{C})| Volume = ∣ A ⋅ ( B × C ) ∣
Example Question Find the volume of the parallelepiped formed by the vectors A ⃗ = ( 1 , − 2 , 3 ) \vec{A} = (1, -2, 3) A = ( 1 , − 2 , 3 ) , B ⃗ = ( 2 , 1 , − 3 ) \vec{B} = (2, 1, -3) B = ( 2 , 1 , − 3 ) , and C ⃗ = ( 3 , 4 , 5 ) \vec{C} = (3, 4, 5) C = ( 3 , 4 , 5 ) .
Solution:
The volume of the parallelepiped is:
Volume = ∣ A ⃗ ⋅ ( B ⃗ × C ⃗ ) ∣ \text{Volume} = |\vec{A} \cdot (\vec{B} \times \vec{C})| Volume = ∣ A ⋅ ( B × C ) ∣
Let's calculate B ⃗ × C ⃗ \vec{B} \times \vec{C} B × C first.
( 2 i + j − 3 k ) × ( 3 i + 4 j + 5 k ) (2i + j - 3k) \times (3i + 4j + 5k) ( 2 i + j − 3 k ) × ( 3 i + 4 j + 5 k )
= ( ( 1 ) ( 5 ) − ( − 3 ) ( 4 ) ) i − ( ( 2 ) ( 5 ) − ( − 3 ) ( 3 ) ) j + ( ( 2 ) ( 4 ) − ( 1 ) ( 3 ) ) k = \left( (1)(5) - (-3)(4) \right)i - \left( (2)(5) - (-3)(3) \right)j + \left( (2)(4) - (1)(3) \right)k = ( ( 1 ) ( 5 ) − ( − 3 ) ( 4 ) ) i − ( ( 2 ) ( 5 ) − ( − 3 ) ( 3 ) ) j + ( ( 2 ) ( 4 ) − ( 1 ) ( 3 ) ) k
= ( 10 + 9 ) i − ( 10 − 9 ) j + ( 8 − 3 ) k = (10 + 9)i - (10 - 9)j + (8 - 3)k = ( 10 + 9 ) i − ( 10 − 9 ) j + ( 8 − 3 ) k
= 19 i − 1 j + 5 k = 19i - 1j + 5k = 19 i − 1 j + 5 k
Now, let's find the dot product of A ⃗ \vec{A} A and B ⃗ × C ⃗ \vec{B} \times \vec{C} B × C .
A ⃗ ⋅ ( 19 i − 1 j + 5 k ) \vec{A} \cdot (19i - 1j + 5k) A ⋅ ( 19 i − 1 j + 5 k )
= ( 1 ) ( 19 ) + ( − 2 ) ( − 1 ) + ( 3 ) ( 5 ) = (1)(19) + (-2)(-1) + (3)(5) = ( 1 ) ( 19 ) + ( − 2 ) ( − 1 ) + ( 3 ) ( 5 )
= 19 + 2 + 15 = 19 + 2 + 15 = 19 + 2 + 15
= 36 = 36 = 36
Therefore, the volume of the parallelepiped is:
∣ Volume ∣ = 36 |\text{Volume}| = 36 ∣ Volume ∣ = 36
Midpoint Formula The coordinates of the midpoint R \text{ }R R of the line segment joining P ( x 1 , y 1 , z 1 ) \text{ }P(x_1, y_1, z_1) P ( x 1 , y 1 , z 1 ) and Q ( x 2 , y 2 , z 2 ) \text{ }Q(x_2, y_2, z_2) Q ( x 2 , y 2 , z 2 ) are:
R ( x 1 + x 2 2 , y 1 + y 2 2 , z 1 + z 2 2 ) \text{ }R\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2}\right) R ( 2 x 1 + x 2 , 2 y 1 + y 2 , 2 z 1 + z 2 )
Example Question Find the midpoint of the line segment joining the points P ( 2 , 3 , − 1 ) \text{ }P(2, 3, -1) P ( 2 , 3 , − 1 ) and Q ( − 1 , 1 , 4 ) \text{ }Q(-1, 1, 4) Q ( − 1 , 1 , 4 ) .
Solution:
The coordinates of the midpoint are:
R ( 2 + ( − 1 ) 2 , 3 + 1 2 , − 1 + 4 2 ) \text{ }R\left(\frac{2 + (-1)}{2}, \frac{3 + 1}{2}, \frac{-1 + 4}{2}\right) R ( 2 2 + ( − 1 ) , 2 3 + 1 , 2 − 1 + 4 )
= ( 1 2 , 4 2 , 3 2 ) \text{ }= \left(\frac{1}{2}, \frac{4}{2}, \frac{3}{2}\right) = ( 2 1 , 2 4 , 2 3 )
= ( 1 2 , 2 , 3 2 ) \text{ }= \left(\frac{1}{2}, 2, \frac{3}{2}\right) = ( 2 1 , 2 , 2 3 )
Therefore, the midpoint of the line segment is ( 1 2 , 2 , 3 2 ) \text{ }\left(\frac{1}{2}, 2, \frac{3}{2}\right) ( 2 1 , 2 , 2 3 ) .
Shortest Distance Between Two Skew Lines The shortest distance d \text{ }d d between two skew lines with direction ratios a 1 = ( a 1 , b 1 , c 1 ) \text{ }\mathbf{a_1} = (a_1, b_1, c_1) a 1 = ( a 1 , b 1 , c 1 ) and a 2 = ( a 2 , b 2 , c 2 ) \text{ }\mathbf{a_2} = (a_2, b_2, c_2) a 2 = ( a 2 , b 2 , c 2 ) and points on the lines r 1 = ( x 1 , y 1 , z 1 ) \text{ }\mathbf{r_1} = (x_1, y_1, z_1) r 1 = ( x 1 , y 1 , z 1 ) and r 2 = ( x 2 , y 2 , z 2 ) \text{ }\mathbf{r_2} = (x_2, y_2, z_2) r 2 = ( x 2 , y 2 , z 2 ) respectively is given by:
d = ∣ ( r 2 − r 1 ) ⋅ ( a 1 × a 2 ) ∣ ∣ a 1 × a 2 ∣ \text{ }d = \frac{|(\mathbf{r_2} - \mathbf{r_1}) \cdot (\mathbf{a_1} \times \mathbf{a_2})|}{|\mathbf{a_1} \times \mathbf{a_2}|} d = ∣ a 1 × a 2 ∣ ∣ ( r 2 − r 1 ) ⋅ ( a 1 × a 2 ) ∣
Example Question Find the shortest distance between the skew lines with direction ratios a 1 ⃗ = ( 1 , − 1 , 2 ) \text{ }\vec{a_1} = (1, -1, 2) a 1 = ( 1 , − 1 , 2 ) and a 2 ⃗ = ( 2 , 3 , − 1 ) \text{ }\vec{a_2} = (2, 3, -1) a 2 = ( 2 , 3 , − 1 ) , and points P ( 1 , 2 , 3 ) \text{ }P(1, 2, 3) P ( 1 , 2 , 3 ) and Q ( − 1 , − 1 , 4 ) \text{ }Q(-1, -1, 4) Q ( − 1 , − 1 , 4 ) lying on the lines respectively.
Solution:
First, we calculate the cross product a 1 × a 2 \text{ }\mathbf{a_1} \times \mathbf{a_2} a 1 × a 2 :
a 1 × a 2 = ( 1 , − 1 , 2 ) × ( 2 , 3 , − 1 ) \text{ }\mathbf{a_1} \times \mathbf{a_2} = (1, -1, 2) \times (2, 3, -1) a 1 × a 2 = ( 1 , − 1 , 2 ) × ( 2 , 3 , − 1 )
= i ( 2 ( − 1 ) − 3 ( 2 ) ) − j ( 1 ( 2 ) − 2 ( − 1 ) ) + k ( 1 ( 3 ) − ( − 1 ) ( − 1 ) ) \text{ }= i(2(-1) - 3(2)) - j(1(2) - 2(-1)) + k(1(3) - (-1)(-1)) = i ( 2 ( − 1 ) − 3 ( 2 )) − j ( 1 ( 2 ) − 2 ( − 1 )) + k ( 1 ( 3 ) − ( − 1 ) ( − 1 ))
= i ( − 2 − 6 ) − j ( 2 + 2 ) + k ( 3 − 1 ) \text{ }= i(-2 - 6) - j(2 + 2) + k(3 - 1) = i ( − 2 − 6 ) − j ( 2 + 2 ) + k ( 3 − 1 )
= i ( − 8 ) − j ( 4 ) + k ( 2 ) \text{ }= i(-8) - j(4) + k(2) = i ( − 8 ) − j ( 4 ) + k ( 2 )
= − 8 i − 4 j + 2 k \text{ }= -8i - 4j + 2k = − 8 i − 4 j + 2 k
Next, we find the vector joining the two points r 2 − r 1 \text{ }\mathbf{r_2} - \mathbf{r_1} r 2 − r 1 :
r 2 − r 1 = ( − 1 − 1 , − 1 − 2 , 4 − 3 ) \text{ }\mathbf{r_2} - \mathbf{r_1} = (-1 - 1, -1 - 2, 4 - 3) r 2 − r 1 = ( − 1 − 1 , − 1 − 2 , 4 − 3 )
= ( − 2 , − 3 , 1 ) \text{ }= (-2, -3, 1) = ( − 2 , − 3 , 1 )
Now, we calculate the dot product of r 2 − r 1 \text{ }\mathbf{r_2} - \mathbf{r_1} r 2 − r 1 and a 1 × a 2 \text{ }\mathbf{a_1} \times \mathbf{a_2} a 1 × a 2 :
( r 2 − r 1 ) ⋅ ( a 1 × a 2 ) = ( − 2 , − 3 , 1 ) ⋅ ( − 8 , − 4 , 2 ) \text{ }(\mathbf{r_2} - \mathbf{r_1}) \cdot (\mathbf{a_1} \times \mathbf{a_2}) = (-2, -3, 1) \cdot (-8, -4, 2) ( r 2 − r 1 ) ⋅ ( a 1 × a 2 ) = ( − 2 , − 3 , 1 ) ⋅ ( − 8 , − 4 , 2 )
= ( − 2 ) ( − 8 ) + ( − 3 ) ( − 4 ) + ( 1 ) ( 2 ) \text{ }= (-2)(-8) + (-3)(-4) + (1)(2) = ( − 2 ) ( − 8 ) + ( − 3 ) ( − 4 ) + ( 1 ) ( 2 )
= 16 + 12 + 2 \text{ }= 16 + 12 + 2 = 16 + 12 + 2
= 30 \text{ }= 30 = 30
Finally, we calculate the magnitude of the cross product a 1 × a 2 \text{ }\mathbf{a_1} \times \mathbf{a_2} a 1 × a 2 :
∣ a 1 × a 2 ∣ = ( − 8 ) 2 + ( − 4 ) 2 + 2 2 \text{ }|\mathbf{a_1} \times \mathbf{a_2}| = \sqrt{(-8)^2 + (-4)^2 + 2^2} ∣ a 1 × a 2 ∣ = ( − 8 ) 2 + ( − 4 ) 2 + 2 2
= 64 + 16 + 4 \text{ }= \sqrt{64 + 16 + 4} = 64 + 16 + 4
= 84 \text{ }= \sqrt{84} = 84
Therefore, the shortest distance between the skew lines is:
d = ∣ 30 ∣ 84 = 30 84 \text{ }d = \frac{|30|}{\sqrt{84}} = \frac{30}{\sqrt{84}} d = 84 ∣30∣ = 84 30
Important Facts → For the vectors to be coplanar, a ⃗ ⋅ ( b ⃗ × c ⃗ ) \vec{a} \cdot (\vec{b} \times \vec{c}) a ⋅ ( b × c ) must be zero:
→ Two vectors a and b are perpendicular if their dot product is zero, i.e., a⋅b=0