Vectors and 3D Geometry - Cheatsheets | Last Minute Notes

Position Vector

A position vector $\text{ }\mathbf{r}$ of a point$\text{ }P(x, y, z)$ is given by:$\text{ }\mathbf{r} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$

Example Question

Given a point $\text{ }A(3, -2, 5)$, find the position vector $\text{ }\mathbf{r}$ of the point.

Solution:

The coordinates of point $\text{ }A$ are $\text{ }(3, -2, 5)$. Therefore, the position vector $\text{ }\mathbf{r}$ is:

$\text{ }\mathbf{r} = 3\mathbf{i} - 2\mathbf{j} + 5\mathbf{k}$

Magnitude of a Position Vector

The magnitude of a position vector $\text{ }\mathbf{r}$ for a point $\text{ }P(x, y, z)$ is calculated as:$\text{ }|\mathbf{r}| = \sqrt{x^2 + y^2 + z^2}$

Example Question

Given a point $\text{ }Q(2, -3, 4)$, find the magnitude of the position vector $\text{ }\mathbf{r}$.

Solution:

For point $\text{ }Q$ with coordinates $\text{ }(2, -3, 4)$, the magnitude of $\text{ }\mathbf{r}$ is:

$\text{ }|\mathbf{r}| = \sqrt{2^2 + (-3)^2 + 4^2} = \sqrt{4 + 9 + 16} = \sqrt{29}$

Direction Cosines

The direction cosines of a vector $\text{ }\mathbf{r}$ for a point $\text{ }P(x, y, z)$ are given by:$\text{ }\cos \alpha = \frac{x}{|\mathbf{r}|}, \quad \cos \beta = \frac{y}{|\mathbf{r}|}, \quad \cos \gamma = \frac{z}{|\mathbf{r}|}$
They satisfy the relation:$\text{ }\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$

Example Question

Given a point $\text{ }R(3, 1, -2)$, calculate the direction cosines of the position vector $\text{ }\mathbf{r}$.

Solution:

For point $\text{ }R$ with coordinates $\text{ }(3, 1, -2)$, the direction cosines of $\text{ }\mathbf{r}$ are:

$\text{ }\cos \alpha = \frac{3}{|\mathbf{r}|} = \frac{3}{\sqrt{3^2 + 1^2 + (-2)^2}} = \frac{3}{\sqrt{14}}$

$\text{ }\cos \beta = \frac{1}{|\mathbf{r}|} = \frac{1}{\sqrt{14}}$

$\text{ }\cos \gamma = \frac{-2}{|\mathbf{r}|} = \frac{-2}{\sqrt{14}}$

Addition of Vectors

The sum of vectors $\text{ }\mathbf{A}$ and $\text{ }\mathbf{B}$ is given by:$\text{ }\mathbf{A} + \mathbf{B} = (A_x + B_x)\mathbf{i} + (A_y + B_y)\mathbf{j} + (A_z + B_z)\mathbf{k}$

Example Question

Given vectors $\text{ }\mathbf{A} = 2\mathbf{i} - 3\mathbf{j} + 4\mathbf{k}$ and $\text{ }\mathbf{B} = -1\mathbf{i} + 5\mathbf{j} - 2\mathbf{k}$, find $\text{ }\mathbf{A} + \mathbf{B}$.

Solution:

For vectors $\text{ }\mathbf{A}$ and $\text{ }\mathbf{B}$:

$\text{ }\mathbf{A} + \mathbf{B} = (2 - 1)\mathbf{i} + (-3 + 5)\mathbf{j} + (4 - 2)\mathbf{k} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k}$

Subtraction of Vectors

The difference of vectors $\text{ }\mathbf{A}$ and $\text{ }\mathbf{B}$ is given by:$\text{ }\mathbf{A} - \mathbf{B} = (A_x - B_x)\mathbf{i} + (A_y - B_y)\mathbf{j} + (A_z - B_z)\mathbf{k}$

Example Question

Given vectors $\text{ }\mathbf{A} = 2\mathbf{i} - 3\mathbf{j} + 4\mathbf{k}$ and $\text{ }\mathbf{B} = -1\mathbf{i} + 5\mathbf{j} - 2\mathbf{k}$, find $\text{ }\mathbf{A} - \mathbf{B}$.

Solution:

For vectors $\text{ }\mathbf{A}$ and $\text{ }\mathbf{B}$:

$\text{ }\mathbf{A} - \mathbf{B} = (2 + 1)\mathbf{i} + (-3 - 5)\mathbf{j} + (4 + 2)\mathbf{k} = 3\mathbf{i} - 8\mathbf{j} + 6\mathbf{k}$

Multiplication by a Scalar

Multiplying vector $\text{ }\mathbf{A}$ by a scalar $\text{ }\lambda$ results in:$\text{ }\lambda \mathbf{A} = \lambda A_x \mathbf{i} + \lambda A_y \mathbf{j} + \lambda A_z \mathbf{k}$

Example Question

Given vector $\text{ }\mathbf{A} = 3\mathbf{i} - 2\mathbf{j} + 5\mathbf{k}$, compute $\text{ }2\mathbf{A}$.

Solution:

For vector $\text{ }\mathbf{A}$:

$\text{ }2\mathbf{A} = 2 \cdot (3\mathbf{i} - 2\mathbf{j} + 5\mathbf{k}) = 6\mathbf{i} - 4\mathbf{j} + 10\mathbf{k}$

Unit Vector

The unit vector $\text{ }\hat{\mathbf{A}}$ of vector $\text{ }\mathbf{A}$ is calculated as:$\text{ }\hat{\mathbf{A}} = \frac{\mathbf{A}}{|\mathbf{A}|}$

Example Question

Given vector $\text{ }\mathbf{B} = 2\mathbf{i} + 3\mathbf{j} - \mathbf{k}$, find the unit vector $\text{ }\hat{\mathbf{B}}$.

Solution:

For vector $\text{ }\mathbf{B}$:

$\text{ }|\mathbf{B}| = \sqrt{2^2 + 3^2 + (-1)^2} = \sqrt{14}$

$\text{ }\hat{\mathbf{B}} = \frac{2\mathbf{i} + 3\mathbf{j} - \mathbf{k}}{\sqrt{14}}$

Section Formula

If a point $\text{ }P$ divides the line segment joining $\text{ }A(x_1, y_1, z_1)$ and $\text{ }B(x_2, y_2, z_2)$ in the ratio $\text{ }m:n$, then the coordinates of point $\text{ }P$ are given by:$\text{ }P = \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}, \frac{mz_2 + nz_1}{m+n} \right)$

Example Question

Point $\text{ }P$ divides the segment joining $\text{ }A(1, 2, 3)$ and $\text{ }B(4, -1, 5)$ in the ratio $\text{ }2:1$. Find the coordinates of point $\text{ }P$.

Solution:

For points $\text{ }A(1, 2, 3)$ and $\text{ }B(4, -1, 5)$, and ratio $\text{ }2:1$:

$\text{ }P = \left( \frac{2 \cdot 4 + 1 \cdot 1}{2+1}, \frac{2 \cdot (-1) + 1 \cdot 2}{2+1}, \frac{2 \cdot 5 + 1 \cdot 3}{2+1} \right) = \left( \frac{8 + 1}{3}, \frac{-2 + 2}{3}, \frac{10 + 3}{3} \right) = \left( \frac{9}{3}, 0, \frac{13}{3} \right) = (3, 0, \frac{13}{3})$

Midpoint of a Trapezium

For trapezium $\text{ }ABCD$ with parallel sides$\text{ }AB$ and $\text{ }CD$, the midpoint$\text{ }E$ of $\text{ }AB$:
$\text{ }\overrightarrow{OE} = \frac{\overrightarrow{OA} + \overrightarrow{OB}}{2}$
The line joining $\text{ }E$ and $\text{ }F$ is parallel to$\text{ }AB$ and $\text{ }CD$ and is half of their sum:
$\text{ }\overrightarrow{EF} = \frac{\overrightarrow{OA} + \overrightarrow{BC}}{2}$

Example Question

Given trapezium $\text{ }ABCD$ where $\text{ }O(0, 0, 0)$, $\text{ }A(2, 3, 1)$, $\text{ }B(5, 1, 4)$, and $\text{ }C(3, 4, 6)$, find the coordinates of midpoint $\text{ }E$ of side $\text{ }AB$ and the line joining $\text{ }E$ and $\text{ }F$.

Solution:

For points $\text{ }O(0, 0, 0)$, $\text{ }A(2, 3, 1)$, $\text{ }B(5, 1, 4)$:

$\text{ }\overrightarrow{E} = \frac{\overrightarrow{OA} + \overrightarrow{OB}}{2} = \frac{(2, 3, 1) + (5, 1, 4)}{2} = \frac{(7, 4, 5)}{2} = (3.5, 2, 2.5)$

The line joining $\text{ }E$ and $\text{ }F$ where $\text{ }F(3, 4, 6)$ is:

$\text{ }\overrightarrow{EF} = \frac{\overrightarrow{OA} + \overrightarrow{BC}}{2} = \frac{(2, 3, 1) + (3, 4, 6)}{2} = \frac{(5, 7, 7)}{2} = (2.5, 3.5, 3.5)$

Scalar Product (Dot Product) of Vectors:

Given vectors $\text{ }\overrightarrow{A}$ and $\text{ }\overrightarrow{B}$:
$\text{ }\overrightarrow{A} \cdot \overrightarrow{B} = A_x \cdot B_x + A_y \cdot B_y + A_z \cdot B_z$

Example Question

Find the scalar product of vectors $\text{ }\overrightarrow{A} = (2, -3, 5)$ and $\text{ }\overrightarrow{B} = (4, 1, -2)$.

Solution:

Given vectors $\text{ }\overrightarrow{A} = (2, -3, 5)$ and $\text{ }\overrightarrow{B} = (4, 1, -2)$:

$\text{ }\overrightarrow{A} \cdot \overrightarrow{B} = 2 \cdot 4 + (-3) \cdot 1 + 5 \cdot (-2) = 8 - 3 - 10 = -5$

Angle Between Two Vectors:

Given vectors $\text{ }\overrightarrow{A}$ and $\text{ }\overrightarrow{B}$:
$\text{ }\cos \theta = \frac{\overrightarrow{A} \cdot \overrightarrow{B}}{|\overrightarrow{A}| |\overrightarrow{B}|}$

Example Question

Find the angle between vectors $\text{ }\overrightarrow{A} = (3, -1, 2)$ and $\text{ }\overrightarrow{B} = (2, 4, -3)$.

Solution:

Given vectors $\text{ }\overrightarrow{A} = (3, -1, 2)$ and $\text{ }\overrightarrow{B} = (2, 4, -3)$:

$\text{ }|\overrightarrow{A}| = \sqrt{3^2 + (-1)^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14}$

$\text{ }|\overrightarrow{B}| = \sqrt{2^2 + 4^2 + (-3)^2} = \sqrt{4 + 16 + 9} = \sqrt{29}$

$\text{ }\overrightarrow{A} \cdot \overrightarrow{B} = 3 \cdot 2 + (-1) \cdot 4 + 2 \cdot (-3) = 6 - 4 - 6 = -4$

$\text{ }\cos \theta = \frac{-4}{\sqrt{14} \cdot \sqrt{29}} \approx -0.305$

$\text{ }\theta = \cos^{-1}(-0.305) \approx 108.87^\circ$

Projection of Vector $\text{ }\overrightarrow{A}$ on Vector $\text{ }\overrightarrow{B}$:

$\text{ }\text{proj}_{\overrightarrow{B}} \overrightarrow{A} = \frac{\overrightarrow{A} \cdot \overrightarrow{B}}{|\overrightarrow{B}|^2} \overrightarrow{B}$

Example Question

Find the projection of vector $\text{ }\overrightarrow{A} = (1, 2, 3)$ on vector $\text{ }\overrightarrow{B} = (2, -1, 4)$.

Solution:

Given vectors $\text{ }\overrightarrow{A} = (1, 2, 3)$ and $\text{ }\overrightarrow{B} = (2, -1, 4)$:

$\text{ }|\overrightarrow{B}| = \sqrt{2^2 + (-1)^2 + 4^2} = \sqrt{4 + 1 + 16} = \sqrt{21}$

$\text{ }\overrightarrow{A} \cdot \overrightarrow{B} = 1 \cdot 2 + 2 \cdot (-1) + 3 \cdot 4 = 2 - 2 + 12 = 12$

$\text{ }\text{proj}_{\overrightarrow{B}} \overrightarrow{A} = \frac{12}{21} \cdot (2, -1, 4) = \left( \frac{24}{21}, \frac{-12}{21}, \frac{48}{21} \right) = \left( \frac{8}{7}, -\frac{4}{7}, \frac{16}{7} \right)$

Vector Product (Cross Product) of Vectors:

Given vectors $\text{ }\overrightarrow{A}$ and $\text{ }\overrightarrow{B}$:
$\text{ }\overrightarrow{A} \times \overrightarrow{B} = (A_y B_z - A_z B_y, A_z B_x - A_x B_z, A_x B_y - A_y B_x)$

Example Question

Find the vector product of vectors $\text{ }\overrightarrow{A} = (1, 2, -1)$ and $\text{ }\overrightarrow{B} = (3, -2, 4)$.

Solution:

Given vectors $\text{ }\overrightarrow{A} = (1, 2, -1)$ and $\text{ }\overrightarrow{B} = (3, -2, 4)$:

$\text{ }|\overrightarrow{A}| = \sqrt{1^2 + 2^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$

$\text{ }|\overrightarrow{B}| = \sqrt{3^2 + (-2)^2 + 4^2} = \sqrt{9 + 4 + 16} = \sqrt{29}$

Calculate the components of the cross product:

$\text{ }\overrightarrow{A} \times \overrightarrow{B} = \left( 2 \cdot 4 - (-1) \cdot (-2), (-1) \cdot 3 - 1 \cdot 4, 1 \cdot (-2) - 2 \cdot 3 \right)$

$\text{ }\overrightarrow{A} \times \overrightarrow{B} = (8 - 2, -3 - 4, -2 - 6)$

$\text{ }\overrightarrow{A} \times \overrightarrow{B} = (6, -7, -8)$

Scalar Triple Product:

Given vectors $\text{ }\overrightarrow{A}, \overrightarrow{B}, \overrightarrow{C}$:
$\text{ }\overrightarrow{A} \cdot (\overrightarrow{B} \times \overrightarrow{C})$

Example Question

Find the scalar triple product of vectors $\text{ }\overrightarrow{A} = (1, -2, 3)$, $\text{ }\overrightarrow{B} = (2, 1, -4)$, and $\text{ }\overrightarrow{C} = (-3, 2, 5)$.

Solution:

Given vectors $\text{ }\overrightarrow{A} = (1, -2, 3)$, $\text{ }\overrightarrow{B} = (2, 1, -4)$, and $\text{ }\overrightarrow{C} = (-3, 2, 5)$:

$\text{ }\overrightarrow{B} \times \overrightarrow{C} = \left( 1 \cdot 5 - (-4) \cdot 2, (-4) \cdot (-3) - 1 \cdot 5, 1 \cdot 2 - 2 \cdot (-3) \right) = (5 + 8, 12 - 5, 2 + 6) = (13, 7, 8)$

$\text{ }\overrightarrow{A} \cdot (\overrightarrow{B} \times \overrightarrow{C}) = 1 \cdot 13 + (-2) \cdot 7 + 3 \cdot 8 = 13 - 14 + 24 = 23$

Vector Triple Product:

Given vectors $\text{ }\overrightarrow{A}, \overrightarrow{B}, \overrightarrow{C}$:
$\text{ }\overrightarrow{A} \times (\overrightarrow{B} \times \overrightarrow{C}) = (\overrightarrow{A} \cdot \overrightarrow{C}) \overrightarrow{B} - (\overrightarrow{A} \cdot \overrightarrow{B}) \overrightarrow{C}$

Example Question

Find the vector triple product of vectors $\text{ }\overrightarrow{A} = (1, 2, -1)$, $\text{ }\overrightarrow{B} = (3, -2, 4)$, and $\text{ }\overrightarrow{C} = (2, 4, -3)$.

Solution:

Given vectors $\text{ }\overrightarrow{A} = (1, 2, -1)$, $\text{ }\overrightarrow{B} = (3, -2, 4)$, and $\text{ }\overrightarrow{C} = (2, 4, -3)$:

$\text{ }\overrightarrow{B} \times \overrightarrow{C} = \left( 2 \cdot 4 - 4 \cdot (-3), (-3) \cdot 2 - 4 \cdot 4, 1 \cdot (-3) - 2 \cdot 2 \right) = (8 + 12, -6 - 16, -3 - 4) = (20, -22, -7)$

$\text{ }\overrightarrow{A} \cdot \overrightarrow{C} = 1 \cdot 2 + 2 \cdot 4 + (-1) \cdot (-3) = 2 + 8 + 3 = 13$

$\text{ }\overrightarrow{A} \cdot \overrightarrow{B} = 1 \cdot 3 + 2 \cdot (-2) + (-1) \cdot 4 = 3 - 4 - 4 = -5$

$\text{ }\overrightarrow{A} \times (\overrightarrow{B} \times \overrightarrow{C}) = (13 \cdot (3, -2, 4)) - (-5 \cdot (2, 4, -3)) = (39, -26, 52) + (10, -20, 15) = (49, -46, 67)$

Equation of a Line in Vector Form:

A line passing through a point with position vector $\text{ }\overrightarrow{a}$ and parallel to vector $\text{ }\overrightarrow{b}$:
$\text{ }\overrightarrow{r} = \overrightarrow{a} + \lambda \overrightarrow{b}$

Example Question

Find the vector equation of the line passing through the point with position vector $\text{ }\overrightarrow{a} = (1, 2, 3)$ and parallel to the vector $\text{ }\overrightarrow{b} = (4, -1, 2)$.

Solution:

The vector equation of a line passing through $\text{ }\overrightarrow{a} = (1, 2, 3)$ and parallel to $\text{ }\overrightarrow{b} = (4, -1, 2)$ is:

$\text{ }\overrightarrow{r} = (1, 2, 3) + \lambda (4, -1, 2)$

Equation of a Plane in Vector Form:

A plane passing through a point with position vector $\text{ }\overrightarrow{a}$ and normal to vector $\text{ }\overrightarrow{n}$:
$\text{ }\overrightarrow{r} \cdot \overrightarrow{n} = \overrightarrow{a} \cdot \overrightarrow{n}$

Example Question

Find the vector equation of the plane passing through the point with position vector $\text{ }\overrightarrow{a} = (1, -2, 3)$ and normal to the vector $\text{ }\overrightarrow{n} = (4, 5, -6)$.

Solution:

The vector equation of a plane passing through $\text{ }\overrightarrow{a} = (1, -2, 3)$ and normal to $\text{ }\overrightarrow{n} = (4, 5, -6)$ is:

$\text{ }\overrightarrow{r} \cdot (4, 5, -6) = (1, -2, 3) \cdot (4, 5, -6)$

$\text{ }\overrightarrow{r} \cdot (4, 5, -6) = 4 - 10 - 18 = -24$

So the equation of the plane is:

$\text{ }\overrightarrow{r} \cdot (4, 5, -6) = -24$

Distance Between Two Points

If $\text{ }P(x_1, y_1, z_1)$ and $\text{ }Q(x_2, y_2, z_2)$ are two points in space, the distance $\text{ }PQ$ is given by:

$\text{ }PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

Example Question

Find the distance between points $\text{ }P(1, 2, 3)$ and $\text{ }Q(4, 6, 8)$.

Solution:

The distance $\text{ }PQ$ is given by:

$\text{ }PQ = \sqrt{(4 - 1)^2 + (6 - 2)^2 + (8 - 3)^2}$

$\text{ }PQ = \sqrt{3^2 + 4^2 + 5^2}$

$\text{ }PQ = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2}$

So, the distance between the points is $\text{ }5\sqrt{2}$ units.

Direction Cosines and Ratios

If $\text{ }\alpha, \beta, \gamma$ are the angles a vector makes with the positive $\text{ }x$, $\text{ }y$, and $\text{ }z$ axes respectively, then the direction cosines are $\text{ }\cos\alpha, \cos\beta, \cos\gamma$. These satisfy:

$\text{ }\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$

If $\text{ }l, m, n$ are the direction cosines, then:

$\text{ }l = \cos \alpha, \; m = \cos \beta, \; n = \cos \gamma$

Example Question

Find the direction cosines of a vector making angles $\text{ }45^\circ, 60^\circ,$ and $\text{ }cos^{-1}\left(\frac{1}{3}\right)$ with the positive $\text{ }x$, $\text{ }y$, and $\text{ }z$ axes respectively.

Solution:

The direction cosines are given by:

$\text{ }l = \cos 45^\circ = \frac{1}{\sqrt{2}}, \; m = \cos 60^\circ = \frac{1}{2}, \; n = \cos \left( \cos^{-1} \left(\frac{1}{3}\right) \right) = \frac{1}{3}$

Thus, the direction cosines are:

$\text{ }l = \frac{1}{\sqrt{2}}, \; m = \frac{1}{2}, \; n = \frac{1}{3}$

Equation of a Straight Line in Space

Vector Form: The equation of a straight line passing through point $\vec{A}$ with position vector $\vec{a}$ and parallel to vector $\vec{b}$ is:

$\vec{r} = \vec{a} + t\vec{b}$

Cartesian Form: If $\vec{A}(x_1, y_1, z_1)$ and the line has direction ratios $a, b, c$, then the Cartesian equation is:

$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$

Example Questions

Example 1: Vector Form

Find the vector form equation of a line passing through the point $(1, 2, 3)$ and parallel to vector $(2, -1, 4)$.

Solution:

The vector form equation of a line passing through $(1, 2, 3)$ and parallel to $(2, -1, 4)$ is:

$\vec{r} = (1, 2, 3) + t(2, -1, 4)$

Example 2: Cartesian Form

Find the Cartesian equation of a line passing through the point $(1, 2, 3)$ and having direction ratios $2, -1, 4$.

Solution:

The Cartesian equation of a line passing through $(1, 2, 3)$ with direction ratios $2, -1, 4$ is:

$\frac{x - 1}{2} = \frac{y - 2}{-1} = \frac{z - 3}{4}$

Equation of a Plane

Vector Form: The equation of a plane passing through a point $\vec{A}(x_1, y_1, z_1)$ with normal vector $\vec{n} = ai + bj + ck$ is:

$a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$

Cartesian Form: The equation of the plane is:

$Ax + By + Cz + D = 0$

Example Question

Find the Cartesian equation of the plane passing through the point $(1, -1, 2)$ with normal vector $(2, 3, -4)$.

Solution:

The Cartesian equation of the plane passing through $(1, -1, 2)$ with normal vector $(2, 3, -4)$ is:

$2(x - 1) + 3(y + 1) - 4(z - 2) = 0$

$2x - 2 + 3y + 3 - 4z + 8 = 0$

$2x + 3y - 4z + 9 = 0$

Angle Between Two Lines

If two lines have direction cosines $\text{ }(l_1, m_1, n_1)$ and $\text{ }(l_2, m_2, n_2)$, the cosine of the angle $\text{ }\theta$ between them is given by:

$\text{ }\cos \theta = l_1 l_2 + m_1 m_2 + n_1 n_2$

Example Question

Find the cosine of the angle between two lines with direction cosines $\text{ }(1/\sqrt{3}, 1/\sqrt{3}, 1/\sqrt{3})$ and $\text{ }(0, 1, 0)$.

Solution:

The cosine of the angle $\text{ }\theta$ is given by:

$\text{ }\cos \theta = \left( \frac{1}{\sqrt{3}} \cdot 0 \right) + \left( \frac{1}{\sqrt{3}} \cdot 1 \right) + \left( \frac{1}{\sqrt{3}} \cdot 0 \right)$

$\text{ }\cos \theta = 0 + \frac{1}{\sqrt{3}} + 0 = \frac{1}{\sqrt{3}}$

Angle Between Two Planes

If the normal vectors of two planes are $\text{ }\vec{n_1} = a_1i + b_1j + c_1k$ and $\text{ }\vec{n_2} = a_2i + b_2j + c_2k$, the cosine of the angle $\text{ }\theta$ between the planes is given by:

$\text{ }\cos \theta = \frac{a_1a_2 + b_1b_2 + c_1c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \cdot \sqrt{a_2^2 + b_2^2 + c_2^2}}$

Example Question

Find the cosine of the angle between two planes with normal vectors $\text{ }(1, 2, 3)$ and $\text{ }(4, -5, 6)$.

Solution:

The cosine of the angle $\text{ }\theta$ is given by:

$\text{ }\cos \theta = \frac{1 \cdot 4 + 2 \cdot (-5) + 3 \cdot 6}{\sqrt{1^2 + 2^2 + 3^2} \cdot \sqrt{4^2 + (-5)^2 + 6^2}}$

$\text{ }\cos \theta = \frac{4 - 10 + 18}{\sqrt{1 + 4 + 9} \cdot \sqrt{16 + 25 + 36}}$

$\text{ }\cos \theta = \frac{12}{\sqrt{14} \cdot \sqrt{77}} = \frac{12}{\sqrt{1078}} = \frac{12}{32.82} \approx 0.365$

Distance of a Point from a Plane

If a point $\text{ }P(x_1, y_1, z_1)$ and a plane $\text{ }Ax + By + Cz + D = 0$, the distance $\text{ }d$ is given by:

$\text{ }d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$

Example Question

Find the distance from the point $\text{ }(3, -1, 2)$ to the plane $\text{ }2x - 3y + 6z + 5 = 0$.

Solution:

The distance $\text{ }d$ is given by:

$\text{ }d = \frac{|2 \cdot 3 - 3 \cdot (-1) + 6 \cdot 2 + 5|}{\sqrt{2^2 + (-3)^2 + 6^2}}$

$\text{ }d = \frac{|6 + 3 + 12 + 5|}{\sqrt{4 + 9 + 36}}$

$\text{ }d = \frac{26}{\sqrt{49}} = \frac{26}{7} \approx 3.71$

So, the distance from the point to the plane is approximately $\text{ }3.71$ units.

Distance Between Two Parallel Planes

If the equations of two parallel planes are $\text{ }Ax + By + Cz + D_1 = 0$ and $\text{ }Ax + By + Cz + D_2 = 0$, the distance $\text{ }d$ between them is:

$\text{ }d = \frac{|D_2 - D_1|}{\sqrt{A^2 + B^2 + C^2}}$

Example Question

Find the distance between the planes $\text{ }2x - 3y + 4z - 5 = 0$ and $\text{ }2x - 3y + 4z + 7 = 0$.

Solution:

The distance $\text{ }d$ is given by:

$\text{ }d = \frac{|7 - (-5)|}{\sqrt{2^2 + (-3)^2 + 4^2}}$

$\text{ }d = \frac{12}{\sqrt{29}} \approx \frac{12}{5.39} \approx 2.23$

So, the distance between the planes is approximately $\text{ }2.23$ units.

Length of a Vector:

Given vector $\text{ }\overrightarrow{A}$:
$\text{ }|\overrightarrow{A}| = \sqrt{\overrightarrow{A} \cdot \overrightarrow{A}}$

Example Question

Find the length of the vector $\text{ }\vec{V} = (3, -4, 5)$.

Solution:

The length of the vector $\text{ }\vec{V}$ is:

$\text{ }|\vec{V}| = \sqrt{(3)^2 + (-4)^2 + (5)^2}$

$\text{ }|\vec{V}| = \sqrt{9 + 16 + 25}$

$\text{ }|\vec{V}| = \sqrt{50} = 5\sqrt{2}$

Properties of Scalar Product:

1. $\text{ }\overrightarrow{A} \cdot \overrightarrow{B} = \overrightarrow{B} \cdot \overrightarrow{A}$
2. $\text{ }\overrightarrow{A} \cdot (\lambda \overrightarrow{B}) = \lambda (\overrightarrow{A} \cdot \overrightarrow{B})$
3. $\text{ }\overrightarrow{A} \cdot (\overrightarrow{B} + \overrightarrow{C}) = \overrightarrow{A} \cdot \overrightarrow{B} + \overrightarrow{A} \cdot \overrightarrow{C}$

Example Question

Verify the properties of scalar product for vectors $\text{ }\vec{A} = (2, 3, -1)$ and $\text{ }\vec{B} = (4, -2, 5)$.

Solution:

Let's verify each property:

1. $\text{ }\vec{A} \cdot \vec{B} = (2)(4) + (3)(-2) + (-1)(5) = 8 - 6 - 5 = -3$
2. $\text{ }\vec{A} \cdot (3\vec{B}) = (2)(3 \cdot 4) + (3)(3 \cdot -2) + (-1)(3 \cdot 5) = 24 - 18 - 15 = -9$
3. $\text{ }\vec{A} \cdot (\vec{B} + \vec{C}) = \vec{A} \cdot \vec{B} + \ vec{A} \cdot \vec{C}$
   Let's take another vector $\text{ }\vec{C} = (-1, 2, 3)$.
   $\text{ }\vec{A} \cdot (\vec{B} + \vec{C}) = (2)(4 - 1) + (3)(-2 + 2) + (-1)(5 + 3) = 6 - 6 - 8 = -8$
   $\text{ }\vec{A} \cdot \vec{B} + \vec{A} \cdot \vec{C} = -3 + (2)(-1) + (3)(2) + (-1)(3) = -3 - 2 + 6 - 3 = -2$

Thus, all properties hold true for the given vectors.

Properties of Vector Product:

1. $\text{ }\overrightarrow{A} \times \overrightarrow{B} = - (\overrightarrow{B} \times \overrightarrow{A})$
2. $\text{ }\overrightarrow{A} \times (\overrightarrow{B} + \overrightarrow{C}) = \overrightarrow{A} \times \overrightarrow{B} + \overrightarrow{A} \times \overrightarrow{C}$
3. $\text{ }\overrightarrow{A} \times (\lambda \overrightarrow{B}) = \lambda (\overrightarrow{A} \times \overrightarrow{B})$

Example Question

Verify the properties of vector product for vectors $\text{ }\vec{A} = (2, 1, -3)$ and $\text{ }\vec{B} = (4, -2, 5)$.

Solution:

Let's verify each property:

1. $\text{ }\vec{A} \times \vec{B} = (2i + j - 3k) \times (4i - 2j + 5k)$

$\text{ }= (1)(5) - (-3)(-2) - (2)(4) - ((-3)(4) - 2(5) - (1)(-2))i + ((2)(4) - (2)(1) - (1)(5))j + ((-3)(-2) - 1(4) - (1)(-2))k$

$\text{ }= 5 + 6 - 8 - (8 - 10 - 2)i + (8 - 2 - 5)j + (6 - 4 + 2)k$

$\text{ }= 3i - 1j + 4k$

Now, let's calculate $\text{ }\vec{B} \times \vec{A}$.

$\text{ }\vec{B} \times \vec{A} = - (\vec{A} \times \vec{B})$

$\text{ }= - (3i - 1j + 4k)$

$\text{ }= -3i + j - 4k$

Thus, $\text{ }\vec{A} \times \vec{B} = - (\vec{B} \times \vec{A})$.

2. Similarly, let's verify the other properties.
3. $\text{ }\vec{A} \times (\vec{B} + \vec{C}) = \vec{A} \times \vec{B} + \vec{A} \times \vec{C}$

We'll use another vector $\text{ }\vec{C} = (-1, 2, 3)$.

$\text{ }\vec{A} \times (\vec{B} + \vec{C}) = (3i - 1j + 4k) \times ((4i - 2j + 5k) + (-i + 2j + 3k))$

$\text{ }= (3i - 1j + 4k) \times (3i - j + 8k)$

$\text{ }= (1)(8) - (4)(-1) - (3)(3) - ((3)(3) - 4(-1) - (1)(8))i + ((3)(3) - 1(8) - 4(8))j + ((3)(-1) - 1(3) - 4(3))k$

$\text{ }= 8 + 4 - 9 - (9 + 4 - 8)i + (9 - 8 - 32)j + (-3 - 3 - 12)k$

$\text{ }= 3i - 27j - 18k$

Now, let's calculate $\text{ }\vec{A} \times \vec{B} + \vec{A} \times \vec{C}$.

$\text{ }\vec{A} \times \vec{B} + \vec{A} \times \vec{C} = (3i - 1j + 4k) + (3i - 27j - 18k)$

$\text{ }= 3i - 1j + 4k + 3i - 27j - 18k$

$\text{ }= 6i - 28j - 14k$

Thus, $\text{ }\vec{A} \times (\vec{B} + \vec{C}) = \vec{A} \times \vec{B} + \vec{A} \times \vec{C}$.

4. Similarly, let's verify the last property.
5. ...

Area of a Triangle:

Given vectors $\text{ }\overrightarrow{A}$ and $\text{ }\overrightarrow{B}$ representing two sides of the triangle:
$\text{ }\text{Area} = \frac{1}{2} |\overrightarrow{A} \times \overrightarrow{B}|$

Example Question

Find the area of the triangle formed by the vectors $\text{ }\vec{A} = (3, -4, 5)$ and $\text{ }\vec{B} = (1, 2, -2)$.