Vectors | IB - Wyatt's Notes

A vector line ($r$) in parametric form takes a scalar parameter ($\gamma$) to produce each point on The line. The line ($\bm{r}(\gamma)$) is a sum of a point on a line ($\bm{a}$ or $\bm{r_0}$) and a Direction vector ($\bm{b}$) scaled by the parameter $\gamma$:

\begin`\{aligned}` \bm{r}(\gamma) = \bm{r_0} + \gamma\bm{b} \end`\{aligned}`

Vectors in cartesian form define a vector by the definition of each coordinate of the unit vector:

\begin`\{aligned}` \frac{x-x_0}{l} = \frac{y-y_0}{m} = \frac{z-z_0}{n} \end`\{aligned}`\begin`\{aligned}` \bm{r}(\gamma) = \bm{r_0} + \gamma\bm{b}\\ \begin`\{pmatrix}` x \\ y \\ z \end`\{pmatrix}` = \begin`\{pmatrix}` x_0 \\ y_0 \\ z_0 \end`\{pmatrix}` + \gamma \begin`\{pmatrix}` l \\ m \\ n \end`\{pmatrix}`\\ x = x_0 + \gamma l, \quad y = y_0 + \gamma m, \quad z = z_0 + \gamma n\\ \gamma = \frac{x-x_0}{l} = \frac{y-y_0}{m} = \frac{z-z_0}{n} \end`\{aligned}`

A vector plane ($\bm{r}$) given in parametric form is defined with the sum of a point ($\bm{a}$ or $\bm{r_0}$) on the plane and two direction vector ($\bm{b}$ and $\bm{c}$) scaled by scalar parameter ($\gamma$ and $\mu$):

\begin`\{aligned}` \bm{r}(\gamma, \mu) = \bm{r_0} + \gamma \bm{b} + \mu \bm{c} \end`\{aligned}`

A vector plane ($\bm{r}$) defined in point-normal form is defined by the dot product between the Normal ($\hat{n}$) and the direction vectors ($\bm{r}-\bm{r_0}$ or $\gamma \bm{b} + \mu \bm{c}$):

\begin`\{aligned}` \left(\bm{r}-\bm{r_0}\right) \cdot \hat{n} = |\bm{r}-\bm{r_0}||\hat{n}|\sin \frac{\pi}{2} = 0\\ \bm{r} \cdot \hat{n} = \bm{r_0} \cdot \hat{n} \end`\{aligned}`

A vector plane ($\bm{r}$) defined in cartesian form is extended from the point-normal form by Expressing normal vector as its individual axis ($x,y,z$):

\begin`\{aligned}` \hat{n} = \begin`\{pmatrix}` a \\ b \\ c \end`\{pmatrix}`\quad \bm{r} = \begin`\{pmatrix}` x \\ y \\ z \end`\{pmatrix}`\quad \bm{r_0} = \begin`\{pmatrix}` x_0 \\ y_0 \\ z_0 \end`\{pmatrix}`\\ a(x-x_0) + b(y-y_0) + c(z-z_0) = 0\\ ax + by + cz = ax_0 + by_0 + cz_0 \end`\{aligned}`

Since $ax_0 + by_0 + cz_0$ produce a constant, the formula booklet uses a single constant $d$ to Represent it:

\begin`\{aligned}` ax + by + cz = d \end`\{aligned}`

A vector $\bm{v}$ in 3D space is represented as a column vector:

\bm{v} = \begin`\{pmatrix}` v_x \\ v_y \\ v_z \end`\{pmatrix}`

Where $v_x$, v_y$$v_z are the components along the x$$y And $z$ axes respectively.

The magnitude of a vector is its length:

$$|`\bm{v}`| = \sqrt{v_x^2 + v_y^2 + v_z^2}$$

A unit vector has magnitude $1$. The unit vector in the direction of $\bm{v}$ is:

$$\hat{v} = \frac{\bm{v}}{|\bm{v}|}$$

A position vector describes the position of a point relative to the origin $O$:

\overrightarrow`\{OP}` = \begin`\{pmatrix}` x \\ y \\ z \end`\{pmatrix}`\bm{a} + \bm{b} = \begin`\{pmatrix}` a_x + b_x \\ a_y + b_y \\ a_z + b_z \end`\{pmatrix}`\quad \bm{a} - \bm{b} = \begin`\{pmatrix}` a_x - b_x \\ a_y - b_y \\ a_z - b_z \end`\{pmatrix}`K\bm{v} = \begin`\{pmatrix}` kv_x \\ kv_y \\ kv_z \end`\{pmatrix}`

The dot product of two vectors produces a scalar:

$$\bm{a} \cdot \bm{b} = |\bm{a}||\bm{b}|\cos\theta$$

Where $\theta$ is the angle between the vectors ($0 \le \theta \le \pi$).

$$\bm{a} \cdot \bm{b} = a_x b_x + a_y b_y + a_z b_z$$$$\cos\theta = \frac{\bm{a} \cdot \bm{b}}{|\bm{a}||\bm{b}|}$$

• $\bm{a} \cdot \bm{a} = |\bm{a}|^2$
• $\bm{a} \cdot \bm{b} = 0$ if and only if $\bm{a} \perp \bm{b}$ (vectors are perpendicular)
• $\bm{a} \cdot \bm{b} = \bm{b} \cdot \bm{a}$ (commutative)
• $\bm{a} \cdot (\bm{b} + \bm{c}) = \bm{a} \cdot \bm{b} + \bm{a} \cdot \bm{c}$ (distributive)

Problem: Find the angle between $\bm{a} = \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}$ and $\bm{b} = \begin{pmatrix} 3 \\ 0 \\ 4 \end{pmatrix}$.

Solution:

$$\bm{a} \cdot \bm{b} = (2)(3) + (1)(0) + (-1)(4) = 6 + 0 - 4 = 2$$ $$|`\bm{a}`| `= \sqrt{4 + 1 + 1} = \sqrt{6}, \quad` |`\bm{b}`| = \sqrt{9 + 0 + 16} = 5$$ $$\cos\theta = \frac{2}{5\sqrt{6}} = \frac{2}{5\sqrt{6}}, \quad \theta = \arccos\left(\frac{2}{5\sqrt{6}}\right) \approx 80.7^{\circ}$$

The cross product of two vectors produces a vector perpendicular to both:

$$|`\bm{a} \times \bm{b}`| = |`\bm{a}`||`\bm{b}`|\sin\theta$$

The direction is given by the right-hand rule.

\bm{a} \times \bm{b} = \begin`\{pmatrix}` a_y b_z - a_z b_y \\ a_z b_x - a_x b_z \\ a_x b_y - a_y b_x \end`\{pmatrix}`

This can be written as a determinant:

\bm{a} \times \bm{b} = \begin`\{vmatrix}` \hat{i} & \hat{j} & \hat{k}\\ a_x & a_y & a_z\\ b_x & b_y & b_z \end`\{vmatrix}`

• $\bm{a} \times \bm{b} = -(\bm{b} \times \bm{a})$ (anti-commutative)
• $\bm{a} \times \bm{a} = \bm{0}$
• $\bm{a} \times \bm{b} = \bm{0}$ if and only if $\bm{a}$ is parallel to $\bm{b}$
• $|\bm{a} \times \bm{b}|$ gives the area of the parallelogram formed by $\bm{a}$ and $\bm{b}$

• Area of a parallelogram: $A = |\bm{a} \times \bm{b}|$
• Area of a triangle: $A = \frac{1}{2}|\bm{a} \times \bm{b}|$
• Volume of a parallelepiped: $V = |\bm{a} \cdot (\bm{b} \times \bm{c})|$ (scalar triple product)

Problem: Find $\bm{a} \times \bm{b}$ where $\bm{a} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$ And $\bm{b} = \begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix}$.

Solution:

\bm{a} \times \bm{b} = \begin`\{pmatrix}` (2)(6) - (3)(5) \\ (3)(4) - (1)(6) \\ (1)(5) - (2)(4) \end`\{pmatrix}` = \begin`\{pmatrix}` 12 - 15 \\ 12 - 6 \\ 5 - 8 \end`\{pmatrix}` = \begin`\{pmatrix}` -3 \\ 6 \\ -3 \end`\{pmatrix}`

To find the intersection of two lines $\bm{r}_1 = \bm{a}_1 + t\bm{b}_1$ and $\bm{r}_2 = \bm{a}_2 + s\bm{b}_2$:

1. Set the parametric equations equal and solve for $t$ and $s$
2. If a solution exists, substitute back to find the intersection point
3. If the direction vectors are parallel ($\bm{b}_1 \times \bm{b}_2 = \bm{0}$), the lines are either parallel or coincident

To find where line $\bm{r} = \bm{a} + t\bm{b}$ intersects plane $\bm{r} \cdot \hat{n} = d$:

1. Substitute $\bm{r} = \bm{a} + t\bm{b}$ into the plane equation: $,$ (\bm{a} + t\bm{b}) \cdot \hat{n} = d $,$
2. Solve for $t$: $,$ t = \frac{d - \bm{a} \cdot \hat{n}}{\bm{b} \cdot \hat{n}} $,$
3. If $\bm{b} \cdot \hat{n} = 0$The line is parallel to the plane (no intersection or lies in the plane)

Problem: Find the intersection of line $\bm{r} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + t\begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}$ With plane $2x - y + z = 5$.

Solution:

Normal vector \hat{n} = \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}$$d = 5.

Point on line: $\bm{a} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}$Direction: $\bm{b} = \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}$.

$$\bm{a} \cdot \hat{n} = (1)(2) + (0)(-1) + (2)(1) = 4$$ $$\bm{b} \cdot \hat{n} = (2)(2) + (1)(-1) + (-1)(1) = 2$$ $$T = \frac{5 - 4}{2} = \frac{1}{2}$$ \bm{r} = \begin`\{pmatrix}` 1 \\ 0 \\ 2 \end`\{pmatrix}` + \frac{1}{2}\begin`\{pmatrix}` 2 \\ 1 \\ -1 \end`\{pmatrix}` = \begin`\{pmatrix}` 2 \\ 0.5 \\ 1.5 \end`\{pmatrix}`

Intersection point: $(2, 0.5, 1.5)$.

The angle $\alpha$ between a line with direction $\bm{b}$ and a plane with normal $\hat{n}$:

$$\sin\alpha = \frac{|\bm{b} \cdot \hat{n}|}{|\bm{b}||\hat{n}|}$$

The angle $\theta$ between two planes with normals $\hat{n}_1$ and $\hat{n}_2$:

$$\cos\theta = \frac{|\hat{n}_1 \cdot \hat{n}_2|}{|\hat{n}_1||\hat{n}_2|}$$

The perpendicular distance from point $P$ with position vector $\bm{p}$ to plane $\bm{r} \cdot \hat{n} = d$:

$$D = \frac{|\bm{p} \cdot \hat{n} - d|}{|\hat{n}|}$$

If $\hat{n}$ is already a unit vector:

$$D = |\bm{p} \cdot \hat{n} - d|$$

The shortest distance between two skew lines $\bm{r}_1 = \bm{a}_1 + t\bm{b}_1$ and $\bm{r}_2 = \bm{a}_2 + s\bm{b}_2$:

$$D = \frac{|(\bm{a}_2 - \bm{a}_1) \cdot (\bm{b}_1 \times \bm{b}_2)|}{|\bm{b}_1 \times \bm{b}_2|}$$

Problem: Find the distance from $P(1, 2, 3)$ to the plane $2x - y + 2z = 4$.

Solution:

Normal vector $\hat{n} = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}$ $|\hat{n}| = \sqrt{4 + 1 + 4} = 3$.

$$\bm{p} \cdot \hat{n} = (1)(2) + (2)(-1) + (3)(2) = 2 - 2 + 6 = 6$$ $$D = \frac{|6 - 4|}{3} = \frac{2}{3}$$

Points A$$B$$C are collinear if and only if $\overrightarrow{AB} = k\overrightarrow{AC}$for Some scalar $k$.

To find the foot of the perpendicular from point $P$ to line $\bm{r} = \bm{a} + t\bm{b}$:

The foot $F$ satisfies:

\overrightarrow`\{PF}` \cdot \bm{b} = 0

Substitute $\bm{F} = \bm{a} + t\bm{b}$ and solve for $t$:

$$(\bm{a} + t\bm{b} - \bm{p}) \cdot \bm{b} = 0 \implies t = \frac{(\bm{p} - \bm{a}) \cdot \bm{b}}{|\bm{b}|^2}$$

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Problem: Find the Cartesian equation of the plane passing through the points $A(1, 2, -1)$ $B(3, 0, 1)$ And $C(0, 1, 3)$.

Solution:

Find two direction vectors in the plane:

\overrightarrow`\{AB}` = \begin`\{pmatrix}` 2 \\ -2 \\ 2 \end`\{pmatrix}`\quad \overrightarrow`\{AC}` = \begin`\{pmatrix}` -1 \\ -1 \\ 4 \end`\{pmatrix}`

Find the normal vector using the cross product:

\hat{n} = \overrightarrow`\{AB}` \times \overrightarrow`\{AC}` = \begin`\{pmatrix}` (-2)(4) - (2)(-1) \\ (2)(-1) - (2)(4) \\ (2)(-1) - (-2)(-1) \end`\{pmatrix}` = \begin`\{pmatrix}` -8 + 2 \\ -2 - 8 \\ -2 - 2 \end`\{pmatrix}` = \begin`\{pmatrix}` -6 \\ -10 \\ -4 \end`\{pmatrix}`

Simplify by dividing by $-2$: $\hat{n} = \begin{pmatrix} 3 \\ 5 \\ 2 \end{pmatrix}$.

Using point $A(1, 2, -1)$ and $\hat{n} \cdot \bm{r} = d$:

$$D = (3)(1) + (5)(2) + (2)(-1) = 3 + 10 - 2 = 11$$

The plane equation is: $3x + 5y + 2z = 11$.

Verification: Check point $B$: $3(3) + 5(0) + 2(1) = 9 + 0 + 2 = 11$. Check point $C$: $3(0) + 5(1) + 2(3) = 0 + 5 + 6 = 11$.

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Problem: Find the acute angle between the planes $\Pi_1: 2x - y + 3z = 7$ and $\Pi_2: x + 4y - z = 3$.

Solution:

Extract the normal vectors:

\hat{n}_1 = \begin`\{pmatrix}` 2 \\ -1 \\ 3 \end`\{pmatrix}`\quad \hat{n}_2 = \begin`\{pmatrix}` 1 \\ 4 \\ -1 \end`\{pmatrix}` $$\hat{n}_1 \cdot \hat{n}_2 = (2)(1) + (-1)(4) + (3)(-1) = 2 - 4 - 3 = -5$$ $$|`\hat{n}_1`| `= \sqrt{4 + 1 + 9} = \sqrt{14}, \quad` |`\hat{n}_2`| = \sqrt{1 + 16 + 1} = \sqrt{18} = 3\sqrt{2}$$ $$\cos\theta = \frac{|\hat{n}_1 \cdot \hat{n}_2|}{|\hat{n}_1||\hat{n}_2|} = \frac{5}{\sqrt{14} \cdot 3\sqrt{2}} = \frac{5}{3\sqrt{28}} = \frac{5}{3 \times 5.292} = 0.315$$ $$\theta = \arccos(0.315) \approx 71.6^{\circ}$$

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Problem: Find the shortest distance from the point $P(3, -1, 2)$ to the plane $x - 2y + 2z = 5$.

Solution:

Normal vector $\hat{n} = \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix}$ $|\hat{n}| = \sqrt{1 + 4 + 4} = 3$.

$$\bm{p} \cdot \hat{n} = (3)(1) + (-1)(-2) + (2)(2) = 3 + 2 + 4 = 9$$ $$D = \frac{|\bm{p} \cdot \hat{n} - d|}{|\hat{n}|} = \frac{|9 - 5|}{3} = \frac{4}{3}$$

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Problem: Find the vector equation of the line of intersection of the planes $\Pi_1: x + y - z = 4$ and $\Pi_2: 2x - y + z = 1$.

Solution:

The direction vector of the line is perpendicular to both normals:

\bm{d} = \hat{n}_1 \times \hat{n}_2 = \begin`\{pmatrix}` 1 \\ 1 \\ -1 \end`\{pmatrix}` \times \begin`\{pmatrix}` 2 \\ -1 \\ 1 \end`\{pmatrix}` = \begin`\{pmatrix}` (1)(1) - (-1)(-1) \\ (-1)(2) - (1)(1) \\ (1)(-1) - (1)(2) \end`\{pmatrix}` = \begin`\{pmatrix}` 0 \\ -3 \\ -3 \end`\{pmatrix}`

Simplify: $\bm{d} = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}$.

Find a point on the line by setting $z = 0$ and solving:

$x + y = 4$ and $2x - y = 1$. Adding: $3x = 5 \implies x = \frac{5}{3}$. Then $y = 4 - \frac{5}{3} = \frac{7}{3}$.

A point on the line is $\left(\frac{5}{3}, \frac{7}{3}, 0\right)$.

The line of intersection is:

\bm{r} = \begin`\{pmatrix}` 5/3 \\ 7/3 \\ 0 \end`\{pmatrix}` + t\begin`\{pmatrix}` 0 \\ 1 \\ 1 \end`\{pmatrix}`

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Vectors are arrows in space — they have both direction and magnitude. The dot product measures how much two arrows point in the same direction: it is maximised when they are parallel and zero when they are perpendicular. The cross product creates a new arrow perpendicular to both originals, and its magnitude equals the area of the parallelogram they span. A plane is defined by a point and a normal direction, and the dot product encodes this: every point on the plane is equidistant from the normal in a specific sense. Intersections happen when you solve simultaneous vector equations, and the distance from a point to a plane is directly how far the point deviates from the plane’s defining equation.

1. Confusing parametric and Cartesian forms. In Cartesian form, each component is equated to a parameter expression. Forgetting to set the ratios equal is a common error.

2. Sign errors in the cross product. The cross product is anti-commutative: $\bm{a} \times \bm{b} = -(\bm{b} \times \bm{a})$. Always double-check the order of vectors.

3. Forgetting the absolute value in distance formulas. The distance from a point to a plane is always non-negative: use $|\bm{p} \cdot \hat{n} - d|$.

4. Assuming a line intersects a plane. Always check that $\bm{b} \cdot \hat{n} \neq 0$ before solving. If $\bm{b} \cdot \hat{n} = 0$ and $\bm{a} \cdot \hat{n} = d$The line lies in the plane. If $\bm{b} \cdot \hat{n} = 0$ and $\bm{a} \cdot \hat{n} \neq d$The line is parallel to the plane.

5. Angle between line and plane vs angle between line and normal. The angle $\alpha$ between a line and a plane satisfies $\sin\alpha = \frac{|\bm{b} \cdot \hat{n}|}{|\bm{b}||\hat{n}|}$. The angle between the line and the normal satisfies $\cos\phi = \frac{|\bm{b} \cdot \hat{n}|}{|\bm{b}||\hat{n}|}$. Note that $\alpha + \phi = 90^\circ$.

6. Assuming skew lines intersect. Two lines in 3D are generally skew (neither parallel nor intersecting). Always verify that a common solution exists for the parameters.

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Question 1

Find the angle between the vectors $\bm{a} = \begin{pmatrix} 1 \\ 3 \\ -2 \end{pmatrix}$ and $\bm{b} = \begin{pmatrix} 4 \\ -1 \\ 1 \end{pmatrix}$.

Answer 1

$\bm{a} \cdot \bm{b} = (1)(4) + (3)(-1) + (-2)(1) = 4 - 3 - 2 = -1$. |\bm{a}| = \sqrt{1 + 9 + 4} = \sqrt{14}$$|\bm{b}| = \sqrt{16 + 1 + 1} = \sqrt{18} = 3\sqrt{2}. $\cos\theta = \frac{-1}{3\sqrt{28}} = \frac{-1}{3 \times 5.292} = -0.0630$. $\theta = \arccos(-0.0630) \approx 93.6^\circ$.

Question 2

Find the Cartesian equation of the plane containing the points P(2, 1, 0)$$Q(1, -1, 3) And $R(4, 0, -1)$.

Answer 2

$\overrightarrow{PQ} = \begin{pmatrix} -1 \\ -2 \\ 3 \end{pmatrix}$ $\overrightarrow{PR} = \begin{pmatrix} 2 \\ -1 \\ -1 \end{pmatrix}$. $\hat{n} = \overrightarrow{PQ} \times \overrightarrow{PR} = \begin{pmatrix} (-2)(-1) - (3)(-1) \\ (3)(2) - (-1)(-1) \\ (-1)(-1) - (-2)(2) \end{pmatrix} = \begin{pmatrix} 5 \\ 5 \\ 5 \end{pmatrix}$. Simplified normal: $\hat{n} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$. $d = (1)(2) + (1)(1) + (1)(0) = 3$. Plane equation: $x + y + z = 3$.

Question 3

Find the point of intersection of the line $\bm{r} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} + t\begin{pmatrix} 3 \\ 2 \\ -1 \end{pmatrix}$ With the plane $2x - y + 2z = 8$.

Answer 3

Substitute into the plane equation: $2(1 + 3t) - (-1 + 2t) + 2(2 - t) = 8$. $2 + 6t + 1 - 2t + 4 - 2t = 8$. $7 + 2t = 8 \implies t = 0.5$. Point: $\begin{pmatrix} 1 + 1.5 \\ -1 + 1 \\ 2 - 0.5 \end{pmatrix} = \begin{pmatrix} 2.5 \\ 0 \\ 1.5 \end{pmatrix}$. Intersection: $(2.5, 0, 1.5)$.

Question 4

Find the acute angle between the planes $x + 2y - 2z = 1$ and $3x - y + z = 4$.

Answer 4

$\hat{n}_1 = \begin{pmatrix} 1 \\ 2 \\ -2 \end{pmatrix}$ $\hat{n}_2 = \begin{pmatrix} 3 \\ -1 \\ 1 \end{pmatrix}$. $\hat{n}_1 \cdot \hat{n}_2 = 3 - 2 - 2 = -1$. $|\hat{n}_1| = \sqrt{1 + 4 + 4} = 3$ $|\hat{n}_2| = \sqrt{9 + 1 + 1} = \sqrt{11}$. $\cos\theta = \frac{|-1|}{3\sqrt{11}} = \frac{1}{9.95} = 0.1005$. $\theta = \arccos(0.1005) \approx 84.2^\circ$.

Question 5

Find the shortest distance from the point $A(2, -3, 1)$ to the plane $2x + y - 2z = 6$.

Answer 5

\hat{n} = \begin{pmatrix} 2 \\ 1 \\ -2 \end{pmatrix}$$|\hat{n}| = \sqrt{4 + 1 + 4} = 3. $\bm{p} \cdot \hat{n} = (2)(2) + (-3)(1) + (1)(-2) = 4 - 3 - 2 = -1$. $D = \frac{|-1 - 6|}{3} = \frac{7}{3}$.

Question 6

Find the shortest distance between the skew lines: $L_1: \bm{r} = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} + s\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}$ And $L_2: \bm{r} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} + t\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$.

Answer 6

$\bm{a}_2 - \bm{a}_1 = \begin{pmatrix} 0 \\ -1 \\ 1 \end{pmatrix}$. $\bm{b}_1 \times \bm{b}_2 = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} \times \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} (0)(0) - (1)(1) \\ (1)(0) - (1)(0) \\ (1)(1) - (0)(0) \end{pmatrix} = \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}$. $|\bm{b}_1 \times \bm{b}_2| = \sqrt{1 + 0 + 1} = \sqrt{2}$. $(\bm{a}_2 - \bm{a}_1) \cdot (\bm{b}_1 \times \bm{b}_2) = (0)(-1) + (-1)(0) + (1)(1) = 1$. $D = \frac{|1|}{\sqrt{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$.

Question 7

Find the foot of the perpendicular from the point $P(4, 1, 3)$ to the line $\bm{r} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} + t\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}$.

Answer 7

$\bm{a} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$ $\bm{b} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}$ $\bm{p} = \begin{pmatrix} 4 \\ 1 \\ 3 \end{pmatrix}$. $\bm{p} - \bm{a} = \begin{pmatrix} 3 \\ -1 \\ 4 \end{pmatrix}$. $(\bm{p} - \bm{a}) \cdot \bm{b} = 6 + 1 + 12 = 19$. $|\bm{b}|^2 = 4 + 1 + 9 = 14$. $t = \frac{19}{14}$. Foot: $\bm{F} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} + \frac{19}{14}\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} = \begin{pmatrix} 1 + 19/7 \\ 2 - 19/14 \\ -1 + 57/14 \end{pmatrix} = \begin{pmatrix} 26/7 \\ 9/14 \\ 43/14 \end{pmatrix}$.

Question 8

Show that the points A(1, 2, 3)$$B(3, 5, 7) And $C(5, 8, 11)$ are collinear.

Answer 8

$\overrightarrow{AB} = \begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix}$ $\overrightarrow{AC} = \begin{pmatrix} 4 \\ 6 \\ 8 \end{pmatrix}$. $\overrightarrow{AC} = 2\overrightarrow{AB}$So $\overrightarrow{AC} = k\overrightarrow{AB}$with $k = 2$. Since one vector is a scalar multiple of the other, the points are collinear.

flowchart TD
    A[2_Vectors] --> B[Key Concepts]
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    C --> F[Design patterns]
    D --> G[Real-world usage]

This topic covers the mathematical techniques and concepts related to vectors, including key theorems, methods, and problem-solving approaches.

Key concepts include:

• coordinate geometry (lines and circles)
• vectors in 2D and 3D
• transformations
• proof and geometric reasoning
• equations of curves

Regular practice with a variety of question types is essential to build fluency and confidence in applying these mathematical techniques.

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[Vectors]	IB	View

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

• Geometry And Trigonometry
• Trigonometry
• Matrices and Transformations