A-Level Further Mathematics – Chapter 6: Further Vectors and Planes Further Pure
Contents
This chapter extends the vector geometry of A-Level Mathematics to three dimensions, introducing the powerful vector product and the full theory of planes. The geometric intuition built here — thinking in terms of normal vectors, perpendicular distances, and parametric descriptions — underpins large areas of further mathematics and mathematical physics.
6.1 Vector Product (Cross Product)
Definition: Vector Product
The vector product (or cross product) of two vectors $\mathbf{a}$ and $\mathbf{b}$ in $\mathbb{R}^3$ is the vector:
$$\mathbf{a} \times \mathbf{b} = |\mathbf{a}||\mathbf{b}|\sin\theta\,\hat{\mathbf{n}}$$where $\theta$ is the angle between $\mathbf{a}$ and $\mathbf{b}$ ($0 \leq \theta \leq \pi$) and $\hat{\mathbf{n}}$ is the unit vector perpendicular to both $\mathbf{a}$ and $\mathbf{b}$ in the direction given by the right-hand rule.
In component form, if $\mathbf{a} = \begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix}b_1\\b_2\\b_3\end{pmatrix}$:
$$\mathbf{a} \times \mathbf{b} = \begin{pmatrix}a_2 b_3 - a_3 b_2 \\ a_3 b_1 - a_1 b_3 \\ a_1 b_2 - a_2 b_1\end{pmatrix}$$Key properties of the vector product:
- Anti-commutativity: $\mathbf{a} \times \mathbf{b} = -(\mathbf{b} \times \mathbf{a})$
- Parallel vectors: $\mathbf{a} \times \mathbf{b} = \mathbf{0}$ if and only if $\mathbf{a}$ and $\mathbf{b}$ are parallel (including $\mathbf{a} = \mathbf{0}$ or $\mathbf{b} = \mathbf{0}$)
- Area of parallelogram: $|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin\theta$ = area of parallelogram with sides $\mathbf{a}$ and $\mathbf{b}$
- Area of triangle: $\frac{1}{2}|\mathbf{a} \times \mathbf{b}|$
Example 6.1.1 — Computing a vector product
Find $\mathbf{a} \times \mathbf{b}$ where $\mathbf{a} = \begin{pmatrix}2\\1\\3\end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix}1\\0\\-1\end{pmatrix}$.
$$\mathbf{a} \times \mathbf{b} = \begin{pmatrix}(1)(-1) - (3)(0) \\ (3)(1) - (2)(-1) \\ (2)(0) - (1)(1)\end{pmatrix} = \begin{pmatrix}-1 - 0 \\ 3 + 2 \\ 0 - 1\end{pmatrix} = \begin{pmatrix}-1 \\ 5 \\ -1\end{pmatrix}$$Verify perpendicularity: $\mathbf{a} \cdot (\mathbf{a} \times \mathbf{b}) = 2(-1) + 1(5) + 3(-1) = -2+5-3 = 0$. ✓ ▮
Example 6.1.2 — Area of a triangle using the vector product
Find the area of the triangle with vertices $A(1,0,2)$, $B(3,1,0)$, $C(2,-1,1)$.
$\overrightarrow{AB} = \begin{pmatrix}2\\1\\-2\end{pmatrix}$, $\overrightarrow{AC} = \begin{pmatrix}1\\-1\\-1\end{pmatrix}$.
$$\overrightarrow{AB} \times \overrightarrow{AC} = \begin{pmatrix}(1)(-1)-(-2)(-1) \\ (-2)(1)-(2)(-1) \\ (2)(-1)-(1)(1)\end{pmatrix} = \begin{pmatrix}-1-2 \\ -2+2 \\ -2-1\end{pmatrix} = \begin{pmatrix}-3\\0\\-3\end{pmatrix}$$ $$\text{Area} = \tfrac{1}{2}|\overrightarrow{AB} \times \overrightarrow{AC}| = \tfrac{1}{2}\sqrt{9+0+9} = \tfrac{1}{2}\sqrt{18} = \tfrac{3\sqrt{2}}{2}$$▮
Example 6.1.3 — Testing for parallel vectors
Determine whether $\mathbf{p} = \begin{pmatrix}2\\-4\\6\end{pmatrix}$ and $\mathbf{q} = \begin{pmatrix}-1\\2\\-3\end{pmatrix}$ are parallel.
$\mathbf{q} = -\frac{1}{2}\mathbf{p}$, so they are parallel. We can confirm: $\mathbf{p} \times \mathbf{q} = \begin{pmatrix}(-4)(-3)-(6)(2)\\(6)(-1)-(2)(-3)\\(2)(2)-(-4)(-1)\end{pmatrix} = \begin{pmatrix}12-12\\-6+6\\4-4\end{pmatrix} = \begin{pmatrix}0\\0\\0\end{pmatrix}$. ▮
Example 6.1.4 — Anti-commutativity
Using $\mathbf{a} = \begin{pmatrix}2\\1\\3\end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix}1\\0\\-1\end{pmatrix}$ from Example 6.1.1:
$$\mathbf{b} \times \mathbf{a} = \begin{pmatrix}(0)(3)-(−1)(1)\\(−1)(2)−(1)(3)\\(1)(1)−(0)(2)\end{pmatrix} = \begin{pmatrix}1\\-5\\1\end{pmatrix} = -(\mathbf{a} \times \mathbf{b})$$This confirms $\mathbf{b} \times \mathbf{a} = -(\mathbf{a} \times \mathbf{b})$. The direction of the cross product depends on the order of the factors. ▮
Exam Tip — The Determinant Mnemonic
The component formula is best remembered as the expansion of a $3\times3$ determinant along the first row:
$$\mathbf{a} \times \mathbf{b} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\a_1&a_2&a_3\\b_1&b_2&b_3\end{vmatrix}$$Expand: $\mathbf{i}(a_2b_3 - a_3b_2) - \mathbf{j}(a_1b_3 - a_3b_1) + \mathbf{k}(a_1b_2 - a_2b_1)$. Note the minus sign on the $\mathbf{j}$ component.
6.2 Equations of a Plane
A plane in $\mathbb{R}^3$ is determined by a point on it and a direction perpendicular to it (the normal vector), or equivalently by a point and two non-parallel direction vectors lying in the plane.
Three Forms of the Equation of a Plane
1. Vector form: $\mathbf{r} = \mathbf{a} + s\mathbf{b} + t\mathbf{c}$, where $\mathbf{a}$ is a position vector of a point on the plane and $\mathbf{b}$, $\mathbf{c}$ are non-parallel direction vectors in the plane. $s, t \in \mathbb{R}$.
2. Scalar product form (normal form): $\mathbf{r} \cdot \mathbf{n} = d$, where $\mathbf{n}$ is a normal vector and $d = \mathbf{a} \cdot \mathbf{n}$ for any point $\mathbf{a}$ on the plane.
3. Cartesian form: $ax + by + cz = d$, where $\mathbf{n} = \begin{pmatrix}a\\b\\c\end{pmatrix}$.
Example 6.2.1 — Writing the normal form from a point and normal
Find the equation of the plane with normal $\mathbf{n} = \begin{pmatrix}2\\1\\-1\end{pmatrix}$ passing through the point $A(3, 0, 1)$.
$d = \mathbf{a} \cdot \mathbf{n} = (3)(2) + (0)(1) + (1)(-1) = 6 - 1 = 5$.
Scalar product form: $\mathbf{r} \cdot \begin{pmatrix}2\\1\\-1\end{pmatrix} = 5$.
Cartesian form: $2x + y - z = 5$. ▮
Example 6.2.2 — Writing the vector form
A plane contains the points $P(1,0,0)$, $Q(0,2,0)$, $R(0,0,3)$. Write the vector equation of the plane.
$\overrightarrow{PQ} = \begin{pmatrix}-1\\2\\0\end{pmatrix}$, $\overrightarrow{PR} = \begin{pmatrix}-1\\0\\3\end{pmatrix}$.
$\mathbf{r} = \begin{pmatrix}1\\0\\0\end{pmatrix} + s\begin{pmatrix}-1\\2\\0\end{pmatrix} + t\begin{pmatrix}-1\\0\\3\end{pmatrix}$, $s, t \in \mathbb{R}$. ▮
Figure 6.1 — A 2D cross-section view of the plane $2x + y - z = 5$. The normal vector $\mathbf{n} = (2, 1, -1)$ is indicated. The plane passes through $(3,0,1)$, $(0,5,0)$, and $(2.5,0,0)$ in various coordinate planes.
6.3 Converting Between Forms
The conversion between vector form and Cartesian form uses the cross product to find the normal from two direction vectors, then the normal form and finally the Cartesian form.
Example 6.3.1 — Vector form to Cartesian form
Convert $\mathbf{r} = \begin{pmatrix}1\\0\\0\end{pmatrix} + s\begin{pmatrix}-1\\2\\0\end{pmatrix} + t\begin{pmatrix}-1\\0\\3\end{pmatrix}$ to Cartesian form.
Step 1 Find the normal $\mathbf{n} = \mathbf{b} \times \mathbf{c}$:
$$\mathbf{n} = \begin{pmatrix}-1\\2\\0\end{pmatrix} \times \begin{pmatrix}-1\\0\\3\end{pmatrix} = \begin{pmatrix}(2)(3)-(0)(0)\\(0)(-1)-(-1)(3)\\(-1)(0)-(2)(-1)\end{pmatrix} = \begin{pmatrix}6\\3\\2\end{pmatrix}$$Step 2 Find $d = \mathbf{a} \cdot \mathbf{n} = (1)(6) + (0)(3) + (0)(2) = 6$.
Step 3 Cartesian form: $6x + 3y + 2z = 6$. ▮
Example 6.3.2 — Cartesian form to vector form
Write $3x - y + 2z = 7$ in vector form.
Step 1 Find a point on the plane by setting two variables to zero. $y = z = 0$: $3x = 7$, $x = 7/3$. Point: $\mathbf{a} = \begin{pmatrix}7/3\\0\\0\end{pmatrix}$.
Step 2 Find two direction vectors lying in the plane. Any vector $\mathbf{b}$ satisfying $\mathbf{n} \cdot \mathbf{b} = 0$, i.e., $3b_1 - b_2 + 2b_3 = 0$.
Choose $b_3 = 0$: $\mathbf{b} = \begin{pmatrix}1\\3\\0\end{pmatrix}$. Choose $b_1 = 0$: $\mathbf{c} = \begin{pmatrix}0\\2\\1\end{pmatrix}$.
$\mathbf{r} = \begin{pmatrix}7/3\\0\\0\end{pmatrix} + s\begin{pmatrix}1\\3\\0\end{pmatrix} + t\begin{pmatrix}0\\2\\1\end{pmatrix}$. ▮
Example 6.3.3 — Normal from three points
Find the Cartesian equation of the plane through $A(2,1,0)$, $B(1,0,3)$, $C(0,2,1)$.
$\overrightarrow{AB} = \begin{pmatrix}-1\\-1\\3\end{pmatrix}$, $\overrightarrow{AC} = \begin{pmatrix}-2\\1\\1\end{pmatrix}$.
$$\mathbf{n} = \overrightarrow{AB} \times \overrightarrow{AC} = \begin{pmatrix}(-1)(1)-(3)(1)\\(3)(-2)-(-1)(1)\\(-1)(1)-(-1)(-2)\end{pmatrix} = \begin{pmatrix}-4\\-5\\-3\end{pmatrix}$$Use point $A$: $d = (-4)(2) + (-5)(1) + (-3)(0) = -13$. So $-4x - 5y - 3z = -13$, i.e., $4x + 5y + 3z = 13$. ▮
6.4 Angles and Distances
- Angle between two planes with normals $\mathbf{n}_1$ and $\mathbf{n}_2$: $\cos\theta = \dfrac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|}$
- Angle between a line (direction $\mathbf{d}$) and a plane (normal $\mathbf{n}$): $\sin\phi = \dfrac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}$
- Distance from point $P$ to plane $\mathbf{r} \cdot \mathbf{n} = d$ (with $|\mathbf{n}| = 1$): $\text{dist} = |\mathbf{p} \cdot \mathbf{n} - d|$
- Distance from point $P(\mathbf{p})$ to plane $ax+by+cz = d$: $\text{dist} = \dfrac{|ap_1 + bp_2 + cp_3 - d|}{\sqrt{a^2+b^2+c^2}}$
Example 6.4.1 — Angle between two planes
Find the acute angle between the planes $2x + y - 2z = 3$ and $x - 2y + 2z = 1$.
$\mathbf{n}_1 = \begin{pmatrix}2\\1\\-2\end{pmatrix}$, $|\mathbf{n}_1| = 3$; $\mathbf{n}_2 = \begin{pmatrix}1\\-2\\2\end{pmatrix}$, $|\mathbf{n}_2| = 3$.
$$\cos\theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|} = \frac{|2 - 2 - 4|}{9} = \frac{4}{9}$$$\theta = \arccos\!\left(\frac{4}{9}\right) \approx 63.6°$. ▮
Example 6.4.2 — Distance from a point to a plane
Find the perpendicular distance from $P(4, 1, -2)$ to the plane $2x + y - z = 5$.
$$\text{dist} = \frac{|2(4) + 1(1) - 1(-2) - 5|}{\sqrt{4 + 1 + 1}} = \frac{|8 + 1 + 2 - 5|}{\sqrt{6}} = \frac{6}{\sqrt{6}} = \sqrt{6}$$▮
Example 6.4.3 — Angle between a line and a plane
Find the angle between the line $\mathbf{r} = \begin{pmatrix}1\\2\\0\end{pmatrix} + t\begin{pmatrix}1\\1\\2\end{pmatrix}$ and the plane $x + 2y - z = 4$.
$\mathbf{d} = \begin{pmatrix}1\\1\\2\end{pmatrix}$, $|\mathbf{d}| = \sqrt{6}$; $\mathbf{n} = \begin{pmatrix}1\\2\\-1\end{pmatrix}$, $|\mathbf{n}| = \sqrt{6}$.
$$\sin\phi = \frac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}||\mathbf{n}|} = \frac{|1 + 2 - 2|}{\sqrt{6}\cdot\sqrt{6}} = \frac{1}{6}$$$\phi = \arcsin\!\left(\frac{1}{6}\right) \approx 9.6°$. ▮
Example 6.4.4 — Distance between parallel planes
Find the distance between the parallel planes $3x + 4z = 10$ and $3x + 4z = -15$.
Pick any point on the first plane, e.g., $P(10/3, 0, 0)$. Find its distance to the second plane:
$$\text{dist} = \frac{|3(10/3) + 4(0) - (-15)|}{\sqrt{9 + 16}} = \frac{|10 + 15|}{5} = \frac{25}{5} = 5$$▮
Exam Tip — Angle Between Line and Plane vs Two Lines
The angle between two lines uses $\cos\theta = |\mathbf{d}_1 \cdot \mathbf{d}_2|/(|\mathbf{d}_1||\mathbf{d}_2|)$. The angle between a line and a plane uses $\sin\phi$ (because the angle is measured from the plane, not from the normal). Do not mix these up — this is one of the most common errors in vector geometry questions.
6.5 Intersection of Lines and Planes
There are three cases to consider: a line meeting a plane, two planes meeting in a line, and three planes meeting at a point, a line, or not at all.
Example 6.5.1 — Line meets a plane (unique intersection)
Find where the line $\mathbf{r} = \begin{pmatrix}1\\2\\1\end{pmatrix} + t\begin{pmatrix}2\\-1\\3\end{pmatrix}$ meets the plane $x + 2y - z = 7$.
Step 1 Substitute the parametric form into the plane equation:
$$(1+2t) + 2(2-t) - (1+3t) = 7$$ $$1 + 2t + 4 - 2t - 1 - 3t = 7 \implies 4 - 3t = 7 \implies t = -1$$Step 2 Find the point: $\mathbf{r} = \begin{pmatrix}1\\2\\1\end{pmatrix} + (-1)\begin{pmatrix}2\\-1\\3\end{pmatrix} = \begin{pmatrix}-1\\3\\-2\end{pmatrix}$.
Check: $(-1) + 2(3) - (-2) = -1 + 6 + 2 = 7$. ✓ ▮
Example 6.5.2 — Line parallel to (or within) a plane
Determine whether the line $\mathbf{r} = \begin{pmatrix}0\\1\\0\end{pmatrix} + t\begin{pmatrix}1\\0\\1\end{pmatrix}$ is parallel to, or lies in, the plane $x - z = 2$.
Check if the direction vector is perpendicular to $\mathbf{n} = \begin{pmatrix}1\\0\\-1\end{pmatrix}$:
$\mathbf{d} \cdot \mathbf{n} = 1(1) + 0(0) + 1(-1) = 0$. So the line is parallel to (or within) the plane.
Check if the base point $(0,1,0)$ satisfies the plane: $0 - 0 = 0 \neq 2$. So the line is parallel to the plane and does not lie in it. ▮
Example 6.5.3 — Intersection of two planes (a line)
Find the line of intersection of the planes $\Pi_1: x + y + z = 6$ and $\Pi_2: 2x - y + z = 3$.
Step 1 Direction vector of the intersection line: $\mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2$.
$$\mathbf{d} = \begin{pmatrix}1\\1\\1\end{pmatrix} \times \begin{pmatrix}2\\-1\\1\end{pmatrix} = \begin{pmatrix}(1)(1)-(1)(-1)\\(1)(2)-(1)(1)\\(1)(-1)-(1)(2)\end{pmatrix} = \begin{pmatrix}2\\1\\-3\end{pmatrix}$$Step 2 Find a point on both planes. Set $z = 0$: $x + y = 6$ and $2x - y = 3$. Adding: $3x = 9$, $x = 3$, $y = 3$.
Line: $\mathbf{r} = \begin{pmatrix}3\\3\\0\end{pmatrix} + t\begin{pmatrix}2\\1\\-3\end{pmatrix}$. ▮
Figure 6.2 — The intersection of two planes projected onto the $xy$-plane. The red line shows the $xy$-trace of $\Pi_1: x+y+z=6$ (setting $z=0$: $x+y=6$) and the blue line the trace of $\Pi_2: 2x-y+z=3$ (setting $z=0$: $2x-y=3$). Their intersection at $(3,3)$ is the base point of the intersection line.
Example 6.5.4 — Three planes: unique point
Find the point common to the three planes $x+y+z=6$, $x-y+2z=5$, $2x+y-z=1$.
Solve the system. Adding equations 1 and 2: $2x + 3z = 11$. Equations 1 and 3: $3x + 2y = 7$. From eq 1 and 2 again, subtract: $2y - z = 1$, so $z = 2y - 1$. Substitute into $2x + 3z = 11$: $2x + 6y - 3 = 11$, $2x + 6y = 14$, $x + 3y = 7$.
Combined with $3x + 2y = 7$: from first, $x = 7 - 3y$; substitute: $3(7-3y) + 2y = 7 \Rightarrow 21 - 7y = 7 \Rightarrow y = 2$. Then $x = 1$, $z = 2(2)-1 = 3$. Point: $(1,2,3)$. ▮
Example 6.5.5 — Three planes: checking consistency
Show that the planes $x+2y+z=4$, $2x+4y+2z=10$, $x+y+z=3$ are inconsistent (no common point).
Plane 2 is $2 \times$ Plane 1 only if the RHS also doubles: $2 \times 4 = 8 \neq 10$. So planes 1 and 2 are parallel and distinct. No point lies on both, hence the three planes have no common solution. ▮
6.6 Geometric Proofs with Vectors
Vectors provide elegant proofs of geometric theorems. The key tools are: the section formula, expressing vectors in terms of a basis, the scalar product for perpendicularity, and the cross product for collinearity checks (parallel vectors).
Key Results for Vector Proofs
- Collinearity: Points $A$, $B$, $C$ are collinear if $\overrightarrow{AB} = \lambda\overrightarrow{AC}$ for some scalar $\lambda$.
- Section formula: If $P$ divides $AB$ in ratio $m:n$, then $\overrightarrow{OP} = \dfrac{n\,\overrightarrow{OA} + m\,\overrightarrow{OB}}{m+n}$.
- Midpoint: $M = \dfrac{1}{2}(\mathbf{a} + \mathbf{b})$.
- Perpendicularity: $\mathbf{a} \perp \mathbf{b} \Leftrightarrow \mathbf{a} \cdot \mathbf{b} = 0$.
Example 6.6.1 — Proving the midpoint theorem
Prove that the line joining the midpoints of two sides of a triangle is parallel to the third side and half its length.
Let the triangle have vertices at $O$, $A$, $B$. Let $M$ be the midpoint of $OA$ and $N$ the midpoint of $OB$.
$$\overrightarrow{OM} = \tfrac{1}{2}\mathbf{a}, \quad \overrightarrow{ON} = \tfrac{1}{2}\mathbf{b}$$ $$\overrightarrow{MN} = \overrightarrow{ON} - \overrightarrow{OM} = \tfrac{1}{2}\mathbf{b} - \tfrac{1}{2}\mathbf{a} = \tfrac{1}{2}(\mathbf{b} - \mathbf{a}) = \tfrac{1}{2}\overrightarrow{AB}$$$\overrightarrow{MN}$ is a scalar multiple of $\overrightarrow{AB}$, so $MN \parallel AB$. Also $|\overrightarrow{MN}| = \frac{1}{2}|\overrightarrow{AB}|$. ▮
Example 6.6.2 — Proving concurrency of medians
Show that the medians of a triangle meet at the centroid $G = \frac{1}{3}(\mathbf{a} + \mathbf{b} + \mathbf{c})$ (where $O$ is the origin and $A$, $B$, $C$ have position vectors $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$).
The median from $A$ to the midpoint $M_{BC} = \frac{1}{2}(\mathbf{b}+\mathbf{c})$ can be parameterised as:
$$\mathbf{r} = \mathbf{a} + t\left(\tfrac{\mathbf{b}+\mathbf{c}}{2} - \mathbf{a}\right) = (1-t)\mathbf{a} + \tfrac{t}{2}\mathbf{b} + \tfrac{t}{2}\mathbf{c}$$At $t = 2/3$: $\mathbf{r} = \frac{1}{3}\mathbf{a} + \frac{1}{3}\mathbf{b} + \frac{1}{3}\mathbf{c} = \frac{1}{3}(\mathbf{a}+\mathbf{b}+\mathbf{c})$.
By symmetry of this expression in $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$, the same point lies on all three medians. ▮
Example 6.6.3 — Perpendicularity proof
Prove that the diagonals of a rhombus are perpendicular.
Let a rhombus have adjacent sides $\mathbf{a}$ and $\mathbf{b}$ from the same vertex, with $|\mathbf{a}| = |\mathbf{b}|$. The diagonals are $\mathbf{a} + \mathbf{b}$ and $\mathbf{a} - \mathbf{b}$.
$$(\mathbf{a}+\mathbf{b}) \cdot (\mathbf{a}-\mathbf{b}) = |\mathbf{a}|^2 - |\mathbf{b}|^2 = 0 \quad (\text{since } |\mathbf{a}| = |\mathbf{b}|)$$The diagonals are perpendicular. ▮
Example 6.6.4 — Ratio result
Point $P$ lies on $AB$ such that $AP:PB = 2:3$. Express $\overrightarrow{OP}$ in terms of $\mathbf{a}$ and $\mathbf{b}$, and find $\overrightarrow{CP}$ in terms of $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$.
$\overrightarrow{OP} = \frac{3\mathbf{a} + 2\mathbf{b}}{5}$ (section formula, ratio $2:3$ from $A$).
$\overrightarrow{CP} = \overrightarrow{OP} - \overrightarrow{OC} = \frac{3\mathbf{a} + 2\mathbf{b}}{5} - \mathbf{c} = \frac{3\mathbf{a} + 2\mathbf{b} - 5\mathbf{c}}{5}$. ▮
Practice Problems
Problem 1
Calculate $\mathbf{a} \times \mathbf{b}$ where $\mathbf{a} = \begin{pmatrix}3\\-1\\2\end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix}1\\4\\-1\end{pmatrix}$. Verify that the result is perpendicular to both $\mathbf{a}$ and $\mathbf{b}$.
Show solution
Check: $\mathbf{a} \cdot (\mathbf{a}\times\mathbf{b}) = 3(-7) + (-1)(5) + 2(13) = -21 - 5 + 26 = 0$. ✓
$\mathbf{b} \cdot (\mathbf{a}\times\mathbf{b}) = 1(-7) + 4(5) + (-1)(13) = -7 + 20 - 13 = 0$. ✓
Problem 2
Find the area of the parallelogram with sides $\mathbf{a} = \begin{pmatrix}1\\0\\3\end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix}2\\1\\-1\end{pmatrix}$.
Show solution
Area $= |\mathbf{a} \times \mathbf{b}| = \sqrt{9 + 49 + 1} = \sqrt{59}$.
Problem 3
Find the Cartesian equation of the plane passing through $A(1,2,3)$, $B(0,1,1)$, and $C(2,0,1)$.
Show solution
$\overrightarrow{AB} = \begin{pmatrix}-1\\-1\\-2\end{pmatrix}$, $\overrightarrow{AC} = \begin{pmatrix}1\\-2\\-2\end{pmatrix}$.
$$\mathbf{n} = \overrightarrow{AB} \times \overrightarrow{AC} = \begin{pmatrix}(-1)(-2)-(-2)(-2)\\(-2)(1)-(-1)(-2)\\(-1)(-2)-(-1)(1)\end{pmatrix} = \begin{pmatrix}2-4\\-2-2\\2+1\end{pmatrix} = \begin{pmatrix}-2\\-4\\3\end{pmatrix}$$$d = (-2)(1) + (-4)(2) + (3)(3) = -2 - 8 + 9 = -1$.
Equation: $-2x - 4y + 3z = -1$, or $2x + 4y - 3z = 1$.
Problem 4
Find the acute angle between the planes $x + 2y + 2z = 5$ and $2x - y + 2z = 3$.
Show solution
$\mathbf{n}_1 = (1,2,2)$, $|\mathbf{n}_1| = 3$; $\mathbf{n}_2 = (2,-1,2)$, $|\mathbf{n}_2| = 3$.
$$\cos\theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{9} = \frac{|2 - 2 + 4|}{9} = \frac{4}{9}$$$\theta = \arccos(4/9) \approx 63.6°$.
Problem 5
Find the perpendicular distance from the point $Q(1,0,3)$ to the plane $4x - y + 8z = 5$.
Show solution
Problem 6
Find the point where the line $\mathbf{r} = \begin{pmatrix}2\\0\\1\end{pmatrix} + t\begin{pmatrix}1\\2\\-1\end{pmatrix}$ meets the plane $3x + y - 2z = 7$.
Show solution
Substituting: $3(2+t) + (2t) - 2(1-t) = 7 \Rightarrow 6 + 3t + 2t - 2 + 2t = 7 \Rightarrow 4 + 7t = 7 \Rightarrow t = 3/7$.
Point: $\mathbf{r} = \begin{pmatrix}2+3/7\\6/7\\1-3/7\end{pmatrix} = \begin{pmatrix}17/7\\6/7\\4/7\end{pmatrix}$.
Problem 7
Find the line of intersection of the planes $x + y - z = 2$ and $x - y + 2z = 1$.
Show solution
Direction: $\mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2 = \begin{pmatrix}1\\1\\-1\end{pmatrix} \times \begin{pmatrix}1\\-1\\2\end{pmatrix} = \begin{pmatrix}(1)(2)-(-1)(-1)\\(-1)(1)-(1)(2)\\(1)(-1)-(1)(1)\end{pmatrix} = \begin{pmatrix}1\\-3\\-2\end{pmatrix}$.
Point (set $z=0$): $x+y=2$, $x-y=1 \Rightarrow x=3/2, y=1/2$.
Line: $\mathbf{r} = \begin{pmatrix}3/2\\1/2\\0\end{pmatrix} + t\begin{pmatrix}1\\-3\\-2\end{pmatrix}$.
Problem 8
Prove, using vectors, that the diagonals of a parallelogram bisect each other.
Show solution
Let the parallelogram have vertices $O$, $A$ (position vector $\mathbf{a}$), $B$ (position vector $\mathbf{b}$), $C = A + B$ (position vector $\mathbf{a}+\mathbf{b}$).
Midpoint of diagonal $OC$: $\frac{1}{2}(\mathbf{0} + \mathbf{a}+\mathbf{b}) = \frac{\mathbf{a}+\mathbf{b}}{2}$.
Midpoint of diagonal $AB$: $\frac{1}{2}(\mathbf{a} + \mathbf{b}) = \frac{\mathbf{a}+\mathbf{b}}{2}$.
The midpoints coincide, so the diagonals bisect each other.
Problem 9
Find the distance between the parallel planes $2x - y + 2z = 6$ and $2x - y + 2z = 12$.
Show solution
A point on the first plane: set $y=z=0$, $x=3$, giving $(3,0,0)$.
Distance to second plane: $\dfrac{|2(3) - 0 + 0 - 12|}{\sqrt{4+1+4}} = \dfrac{|6-12|}{3} = \dfrac{6}{3} = 2$.
Problem 10
Point $P$ divides $\overrightarrow{AB}$ such that $AP:PB = 1:2$. Given $A$ has position vector $\begin{pmatrix}4\\1\\-1\end{pmatrix}$ and $B$ has position vector $\begin{pmatrix}1\\4\\2\end{pmatrix}$, find the position vector of $P$ and the distance $|OP|$.
Show solution
Section formula (ratio $1:2$ from $A$): $\overrightarrow{OP} = \dfrac{2\mathbf{a} + 1\mathbf{b}}{3} = \dfrac{1}{3}\left[\begin{pmatrix}8\\2\\-2\end{pmatrix}+\begin{pmatrix}1\\4\\2\end{pmatrix}\right] = \dfrac{1}{3}\begin{pmatrix}9\\6\\0\end{pmatrix} = \begin{pmatrix}3\\2\\0\end{pmatrix}$.
$|OP| = \sqrt{9+4+0} = \sqrt{13}$.