H2 Math Vectors: Study Guide

JC · H2 Mathematics · Study guide

Work through ratios, vector algebra, equations and relationships of lines and planes, angles, shortest distances, perpendicular feet, reflections and geometrical loci.

Concepts · worked methods · graphs10 min read · practice linked below

Identify points, directions and normals

A position vector locates a point. Subtract two position vectors to obtain a direction. A scalar product tests perpendicularity; a vector product produces a vector perpendicular to both inputs. Start with the geometrical condition, then translate it into an equation.

Ratio, scalar product and vector product

Ratios lead to an intersection and an area

Let A and B have non-parallel position vectors $\mathbf a$ and $\mathbf b$ . C lies on OA with OC:CA = 2:1. D lies on OB produced with OD:BD = 3:2. Find the intersection E of BC and AD, and express the area of triangle CDE in terms of $|\mathbf a\times\mathbf b|$ .

For C use the internal ratio. For D, OB is the difference between OD and BD, so OD is three times OB.

\begin{aligned}\mathbf c&=\frac23\mathbf a,\\\mathbf d&=3\mathbf b.\end{aligned}

Use different line parameters and equate the independent a and b coefficients:

\begin{aligned}\mathbf e&=\mathbf b+\lambda\left(\frac23\mathbf a-\mathbf b\right),\\&=\mathbf a+\mu(3\mathbf b-\mathbf a),\\\frac23\lambda&=1-\mu,\\1-\lambda&=3\mu,\\\lambda&=2,\qquad\mu=-\frac13,\\\mathbf e&=\frac43\mathbf a-\mathbf b.\end{aligned}

The negative μ places E beyond A away from D; an intersection need not lie inside both segments. Use two sides of triangle CDE:

\begin{aligned}\overrightarrow{CD}&=3\mathbf b-\frac23\mathbf a,\\\overrightarrow{CE}&=\frac23\mathbf a-\mathbf b,\\\overrightarrow{CD}\times\overrightarrow{CE}&=\left(\frac23-2\right)\mathbf a\times\mathbf b,\\&=-\frac43\mathbf a\times\mathbf b,\\\text{Area of CDE}&=\frac12\left|\overrightarrow{CD}\times\overrightarrow{CE}\right|,\\&=\frac23|\mathbf a\times\mathbf b|.\end{aligned}

Taking the magnitude makes the area positive. Reversing the order of the cross product changes its sign, not the area.

A vector equation determines a scalar parameter

\begin{aligned}\mathbf a\cdot\mathbf c&=1(\sqrt3)\cos\frac\pi6=\frac32,\\|3\mathbf a+\mathbf c|^2&=|k\mathbf b|^2,\\9|\mathbf a|^2+6\mathbf a\cdot\mathbf c&\\{}+|\mathbf c|^2&=k^2|\mathbf b|^2,\\9+6\left(\frac32\right)+3&=9k^2,\\k^2&=\frac73,\\k&=\pm\sqrt{\frac73}.\end{aligned}

The norm equation gives two signs. Neither is excluded by the stated information.

Rearrange a vector-product identity

Let $\mathbf a\ne\mathbf0$ and $\mathbf a\times3\mathbf b=-2\mathbf c\times\mathbf a$ . Show that $3\mathbf b-2\mathbf c$ is a scalar multiple of $\mathbf a$ .

\begin{aligned}\mathbf c\times\mathbf a&=-\mathbf a\times\mathbf c,\\\mathbf a\times3\mathbf b&=2\mathbf a\times\mathbf c,\\\mathbf a\times(3\mathbf b-2\mathbf c)&=\mathbf0,\\3\mathbf b-2\mathbf c&=\lambda\mathbf a.\end{aligned}

A zero cross product means that the second vector is parallel to the non-zero a, or is zero; both are included by a scalar λ.

Form equations of lines and planes

Use two directions in a plane

Let $A=(1,0,1)$ , $B=(3,1,1)$ , and $C=(1,2,2)$ . Find a line through A and B, the plane ABC, and the area of triangle ABC.

\begin{aligned}\overrightarrow{AB}&=(2,1,0),\\\overrightarrow{AC}&=(0,2,1),\\\ell:\ \mathbf r&=(1,0,1)+\lambda(2,1,0).\end{aligned}

For the line, a Cartesian form is $(x-1)/2=y$ , $z=1$ . Do not divide by the zero z-component. A normal to the plane is

\begin{aligned}\mathbf n&=(2,1,0)\times(0,2,1),\\&=(1,-2,4).\end{aligned}

\begin{aligned}\mathbf r\cdot\mathbf n&=(1,0,1)\cdot(1,-2,4),\\&=5.\end{aligned}

\begin{aligned}\pi:\ x-2y+4z&=5.\end{aligned}

The parametric plane uses both independent directions:

\begin{aligned}\mathbf r&=(1,0,1)+s(2,1,0)\\&\quad+t(0,2,1),\quad s,t\in\mathbb R.\end{aligned}

\begin{aligned}\text{Area of ABC}&=\frac12|(1,-2,4)|,\\&=\frac{\sqrt{21}}2.\end{aligned}

Substitute all three points into the Cartesian plane as a check. A direction vector and a normal vector have different roles.

Non-parallel lines may be skew

Consider $\ell_1:\mathbf r=(1,0,1)+\lambda(1,1,-2)$ and $\ell_2:\mathbf r=(1,2,-1)+\mu(7,5,1)$ . Determine their relationship and their shortest distance.

The direction vectors are not scalar multiples. To check whether the lines meet, equate the three coordinates:

\begin{aligned}\lambda&=7\mu,\\\lambda&=2+5\mu,\\1-2\lambda&=-1+\mu.\end{aligned}

The first two equations give μ = 1 and λ = 7. The third would require −13 = 0, which is false. The lines are non-parallel and non-intersecting, so they are skew.

Their shortest joining segment is perpendicular to both directions. Project a vector between the lines onto that common normal:

\begin{aligned}\mathbf n&=(1,1,-2)\times(7,5,1),\\&=(11,-15,-2),\\\overrightarrow{AB}&=(1,2,-1)-(1,0,1),\\&=(0,2,-2),\\d&=\frac{|\overrightarrow{AB}\cdot\mathbf n|}{|\mathbf n|},\\&=\frac{|-30+4|}{\sqrt{121+225+4}},\\&=\frac{26}{5\sqrt{14}}.\end{aligned}

This projection gives a distance, not a point on either line. A zero dot product between direction vectors alone would not establish that the lines meet.

Intersections, angles and distances

Two intersecting planes and a line–plane angle

Use $\pi:x-2y+4z=5$ and $\sigma:x+y=2$ . Find their intersection line, their acute angle, and the acute angle between π and a line with direction $(1,0,1)$ .

The intersection direction is perpendicular to both normals. Find one common point by setting z = 0.

\begin{aligned}\mathbf d&=(1,-2,4)\times(1,1,0),\\&=(-4,4,3),\\x-2y&=5,\qquad x+y=2,\\(x,y,z)&=(3,-1,0),\\\pi\cap\sigma:\ \mathbf r&=(3,-1,0)+t(-4,4,3).\end{aligned}

For the acute angle α between planes, use the acute angle between their normals:

\begin{aligned}\cos\alpha&=\frac{|(1,-2,4)\cdot(1,1,0)|}{\sqrt{21}\sqrt2},\\&=\frac1{\sqrt{42}},\\\alpha&\approx81.124^\circ.\end{aligned}

For the angle β between a line and a plane, use sine because the normal is perpendicular to the plane:

\begin{aligned}\sin\beta&=\frac{|(1,0,1)\cdot(1,-2,4)|}{\sqrt2\sqrt{21}},\\&=\frac5{\sqrt{42}},\\\beta&\approx50.490^\circ.\end{aligned}

The modulus selects the acute angle. For the angle between directed vectors, retain the signed dot product.

A line parallel to a plane may lie inside it

The line is $\ell:\mathbf r=(1,0,1)+\lambda(2,1,0)$ . Determine its relationship with $\rho:x-2y+4z=q$ .

\begin{aligned}(2,1,0)\cdot(1,-2,4)&=2-2=0.\end{aligned}

\begin{aligned}x-2y+4z&=(1+2\lambda)-2\lambda+4,\\&=5.\end{aligned}

If q = 5, every point on the line satisfies the plane, so the line is contained in it. If q ≠ 5, no point satisfies it, so the line is parallel and separate.

For comparison, the line $\mathbf r=\mu(1,0,1)$ intersects $\pi:x-2y+4z=5$ :

\begin{aligned}\mu+4\mu&=5,\\\mu&=1,\\\text{intersection}&=(1,0,1).\end{aligned}

A unique parameter means one intersection; an identity means the line is in the plane; an inconsistent equation means no intersection.

Three planes: one point, a line, or no common point

Two planes are $x+y+z=1$ , $x-y+z=3$ . A third is $x+ay+bz=c$ . Classify their common intersection in terms of a, b, c.

\begin{aligned}2y&=-2,\qquad y=-1,\\x+z&=2,\\\mathbf r&=(2,-1,0)+s(-1,0,1).\end{aligned}

Substitute this entire line into the third plane:

\begin{aligned}(2-s)-a+bs&=c,\\(b-1)s&=c-2+a.\end{aligned}

If $b\ne1$ , there is one value of s, hence one common point. If $b=1$ and $c=2-a$ , every s works, hence all three planes contain the line. If $b=1$ and $c\ne2-a$ , the equation is inconsistent and there is no common point.

For example, $(a,b,c)=(2,2,0)$ gives the single point $(2,-1,0)$ ; $(2,1,0)$ gives a common line; $(2,1,1)$ gives no common point.

Foot of perpendicular and reflection

From a point to a line: locate the foot first

Let $A=(3,1,3)$ , and let the line OB have direction $\mathbf b=(5,-4,3)$ through the origin. Find the perpendicular foot H, the distance from A to the line, and the reflection A′ in the line.

H lies on the line, so write its position vector with one parameter. The joining segment AH must be perpendicular to the line direction.

\begin{aligned}\mathbf h&=\lambda(5,-4,3),\\(\mathbf h-\mathbf a)\cdot\mathbf b&=0,\\0&=5(5\lambda-3)\\&\quad-4(-4\lambda-1)\\&\quad+3(3\lambda-3),\\0&=50\lambda-20,\\\lambda&=\frac25,\\\mathbf h&=\left(2,-\frac85,\frac65\right).\end{aligned}

\begin{aligned}\overrightarrow{AH}&=\left(-1,-\frac{13}5,-\frac95\right),\\d&=\sqrt{1+\frac{169}{25}+\frac{81}{25}},\\&=\sqrt{11},\\\mathbf a'&=2\mathbf h-\mathbf a,\\&=\left(1,-\frac{21}5,-\frac35\right).\end{aligned}

H is the midpoint of AA′. As a separate projection check:

\begin{aligned}\text{projection length}&=\frac{|\mathbf a\cdot\mathbf b|}{|\mathbf b|},\\&=\frac{20}{\sqrt{50}}=2\sqrt2,\\\text{projection vector}&=\frac{\mathbf a\cdot\mathbf b}{|\mathbf b|^2}\mathbf b,\\&=\frac25\mathbf b=\mathbf h.\end{aligned}

A projection length is a non-negative number. A projection vector includes direction. They are not interchangeable.

From a point to a plane: move along the normal

Find the perpendicular foot and reflection of $P=(0,0,0)$ in the plane $x+y=2$ , and its distance from the plane.

\begin{aligned}\mathbf n&=(1,1,0),\\\mathbf h&=(0,0,0)+\lambda(1,1,0),\\h_x+h_y&=2,\\2\lambda&=2,\qquad\lambda=1,\\H&=(1,1,0),\\P'&=2H-P=(2,2,0),\\PH&=\sqrt{1^2+1^2}=\sqrt2.\end{aligned}

Both conditions are visible: H lies on the plane, and PH is parallel to its normal. For a variable P = (u,0,0), the same working gives

\begin{aligned}\mathbf h&=(u+\lambda,\lambda,0),\\u+2\lambda&=2,\\\lambda&=\frac{2-u}{2},\\H&=\left(\frac{u+2}{2},\frac{2-u}{2},0\right),\\P'&=(2,2-u,0),\\d&=\frac{|u-2|}{\sqrt2}.\end{aligned}

This is the family shown in the adjacent graph.

Move the point and keep both reflection conditions

u = 0.0: H = (1.00, 1.00, 0), P′ = (2.00, 2.00, 0). Distance = 1.414. H is the midpoint and lies on x + y = 2.

Distances between parallel planes

Use the same normal before comparing constants

Find the distance between $x+y=2$ and $2x+2y=10$ . Then find the planes parallel to $x+y=2$ at distance 2 from it.

Divide the second equation by 2 before subtracting constants:

\begin{aligned}2x+2y=10&\iff x+y=5,\\d&=\frac{|5-2|}{\sqrt{1^2+1^2}},\\&=\frac3{\sqrt2}.\end{aligned}

For an unknown plane x + y = k at distance 2:

\begin{aligned}\frac{|k-2|}{\sqrt2}&=2,\\k-2&=\pm2\sqrt2,\\x+y&=2+2\sqrt2\\\text{or}\quad x+y&=2-2\sqrt2.\end{aligned}

There are two offset planes, one on each side. If finding a point-to-plane distance, the same normal projection gives |p · n − k| / |n|.

Reflect an entire line or plane

Reflect enough points to determine the image

Reflect the z-axis and the plane $x=0$ in $x+y=2$ . The point mapping, obtained from the foot method, is $(x,y,z)\mapsto(2-y,2-x,z)$ .

Two distinct points determine a line:

\begin{aligned}(0,0,0)&\mapsto(2,2,0),\\(0,0,1)&\mapsto(2,2,1).\end{aligned}

The reflected z-axis is

\begin{aligned}\mathbf r&=(2,2,0)+t(0,0,1).\end{aligned}

Choose three non-collinear points in x = 0 for the plane:

\begin{aligned}(0,0,0)&\mapsto(2,2,0),\\(0,1,0)&\mapsto(1,2,0),\\(0,0,1)&\mapsto(2,2,1).\end{aligned}

The image points are non-collinear and all have y = 2, so the reflected plane is y = 2. Their common intersection with the reflecting plane, x = 0, y = 2, is fixed.

Geometrical meanings: translate the vector condition

A dot-product locus is a sphere

The variable point R has position vector $\mathbf r=(x,y,z)$ . Let $A=(-1,0,0)$ , $B=(3,0,0)$ . Describe the locus $(\mathbf r-\mathbf a)\cdot(\mathbf r-\mathbf b)=0$ .

\begin{aligned}(x+1)(x-3)+y^2+z^2&=0,\\x^2-2x-3+y^2+z^2&=0,\\(x-1)^2+y^2+z^2&=4.\end{aligned}

The locus is the sphere with centre (1,0,0) and radius 2. Geometrically, RA and RB are perpendicular away from the endpoints, and AB is a diameter. The algebraic equation also includes A and B.

Compare these related conditions: $(\mathbf r-\mathbf a)\times\mathbf d=\mathbf0$ , with non-zero d, gives a line through A parallel to d. The condition $(\mathbf r-\mathbf a)\cdot\mathbf n=0$ , with non-zero n, gives a plane through A normal to n.

Parameters in scalar and vector products

Collinearity and a greatest projection

Let $\mathbf p=(\cos t,\sin t,-1)$ , $\mathbf q=(\cos2t,-\sin2t,\alpha)$ , with $0<t<\pi$ . Find the parameter values for O, P, Q to be collinear. Then, for α = 1/2, find the greatest projection length of q onto p.

\begin{aligned}\mathbf p\times\mathbf q&=(\alpha\sin t-\sin2t,\\&\quad-\cos2t-\alpha\cos t,\\&\quad-\sin3t).\end{aligned}

For collinearity every component must be zero. The last one gives sin 3t = 0:

\begin{aligned}t&=\frac\pi3\quad\text{or}\quad\frac{2\pi}3.\end{aligned}

Use the first component to obtain α and check the second component as well:

\begin{aligned}t=\frac\pi3&\implies\alpha=1,\\t=\frac{2\pi}3&\implies\alpha=-1.\end{aligned}

For the projection question, set α = 1/2 and use a scalar product:

\begin{aligned}\mathbf p\cdot\mathbf q&=\cos t\cos2t-\sin t\sin2t-\frac12,\\&=\cos3t-\frac12,\\|\mathbf p|&=\sqrt2,\\L&=\frac{|\cos3t-1/2|}{\sqrt2}.\end{aligned}

Since $-1\leq\cos3t\leq1$ , the largest numerator is 3/2, attained at $t=\pi/3$ , which is allowed. Hence $L_{\max}=3/(2\sqrt2)$ . The negative dot product at this point still gives a positive length.

Before you finish

Use different parameters for different lines. Check all three coordinates at an intersection. Verify a foot is on its line or plane and that the joining vector is perpendicular. For reflection, check the midpoint. Distinguish an acute angle, a directed-vector angle, a distance and a projection vector.

Vectors practice questions and worked solutions