Further Pure | Chapter 7: Vectors

Curriculum Outcome:

• use the equation of a plane in any of the forms ax + by + cz = d or r.n = p or r = a + λb + µc, and convert equations of planes from one form to another as necessary in solving problems;
• recall that the vector product a × b of two vectors can be expressed either as |a| |b| sin θ nˆ , where nˆ is a unit vector, or in component form as (a2b3 – a3b2) i + (a3b1 – a1b3) j + (a1b2 – a2b1) k;
• use equations of lines and planes, together with scalar and vector products where appropriate, to solve problems concerning distances, angles and intersections, including: determining whether a line lies in a plane, is parallel to a plane or intersects a plane, and finding the point of intersection of a line and a plane when it exists, finding the perpendicular distance from a point to a plane, finding the angle between a line and a plane, and the angle between two planes, finding an equation for the line of intersection of two planes, calculating the shortest distance between two skew lines. finding an equation for the common perpendicular to two skew lines

We shall not tarry on topics you should have learnt in the A Levels Mathematics Course. The majority of the questions on the topic have been simply about the applications of different formulae, that too, mostly straightforward. This is in my reckoning the easiest chapter in your syllabus.

Equation of a Plane

If we know a vector perpendicular to the plane, n, we wish to find an equation to, and further a point on the plane, a, and the general point r, the equation is:

$r \cdot \vec n = a \cdot \vec n$

Or if we use matrices to represent the vectors:

$\left( {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}} {{n_1}}\\ {{n_2}}\\ {{n_3}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} a\\ b\\ c \end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}} {{n_1}}\\ {{n_2}}\\ {{n_3}} \end{array}} \right)$

An alternative method of finding the vector equation would require non-collinear points (A, B, C) on the plane. If we have the position vectors of such three points, i.e.

 $OA = a;\quad OB = b;\quad OC = c$

Then the equation of the plane is:

$r = \vec a + m\left( {\vec b - \vec a} \right) + n\left( {\vec c - \vec a} \right)$

Ex. Find the vector equation of the plane containing the points $a = j + k;\quad b = i + j;\quad c = i + k$ and convert it to the cartesian form.

Sol. Given in the question is:

$a = j + k;\quad b = i + j;\quad c = i + k$

The plane then becomes:

$\begin{array}{l} r = j + k + m\left( {\left( {i + j} \right) - \left( {j + k} \right)} \right) + n\left( {\left( {i + k} \right) - \left( {j + k} \right)} \right)\\ r = j + k + m\left( {i - k} \right) + n\left( {i - j} \right) \end{array}$

Comparing it with the cartesian form, which is:

$r = xi + yj + zk$

I get:

$x = m + n;\quad y = 1 - n;\quad z = 1 - m$

Substituting in the values of m and n, into the equation involving x, I get:

$\begin{array}{l} x = \left( {1 - z} \right) + \left( {1 - y} \right)\\ x + y + z = 2 \end{array}$

Vector Product (Cross Product)

In the A Levels Mathematics Course, you must have met the scalar product or the dot product, which is one manner of multiplying two vector. The multiplication of vectors actually involves a couple, one moiety of which is the Vector Product. Perhaps so named, for it produces a vector as a result, in contradiction to the scalar product which produces a scalar quantity.

We define the vector product of the above two vectors, as:

$\vec a \times \vec b = \left| a \right|\left| b \right|\sin \theta {\kern 1pt} \hat n$

The right hand side has the product of the moduli of the vectors, the sine of the acute angle between the vectors, and a unit vector in the direction determined by the right hand rule. For $a \times b$ one curls the fingers from a to b, and the direction of the unit vector is in the direction of the thumb. For $b \times a$ we have therefore:

$\vec b \times \vec a = \left| a \right|\left| b \right|\sin \theta {\kern 1pt} \left( { - \hat n} \right) =  - \left| a \right|\left| b \right|\sin \theta {\kern 1pt} \hat n$

We have that:

$\left( {\vec b \times \vec a} \right) + \left( {\vec a \times \vec b} \right) = 0$

It may also be surmised that the vector product is not, in general, commutative, i.e.

$\left( {\vec b \times \vec a} \right) \ne \left( {\vec a \times \vec b} \right)$

Also note:

$a \times a = 0$

A simple example to give you feel of the matter.

Ex. Find the cross product of the vectors $a = 2i;\quad b = 5j$.

Sol.

As the vectors lie on different axes the angles between them is 90 degrees. the relevant unit vector is the z axis. The Vector Product is then:

$a \times b = \left( 2 \right)\left( 5 \right)\sin \left( {90} \right)\left( k \right) = 10k$

Now let us detail the method you must have met in before. If the vectors are given in the matrix form, or can be brought into it, the cross product is defined for two vectors as:

$\left( {\begin{array}{*{20}{c}} {{a_1}}\\ {{a_2}}\\ {{a_3}} \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} {{b_1}}\\ {{b_2}}\\ {{b_3}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {{a_2}{b_3} - {a_3}{b_2}}\\ {{a_3}{b_1} - {a_1}{b_3}}\\ {{a_1}{b_2} - {a_2}{b_1}} \end{array}} \right)$

You can either learn this, as it is, or you may learn the determinant approach. For the two vectors considered above, we produce a matrix of the form below and find the determinant of it. If  matrices and determinants are notions, oblivious to you, you may completely ignore this part and move on. It will be revisited in great detail when we begin the chapter linear algebra.

$\left( {\begin{array}{*{20}{c}} i&j&k\\ {{a_1}}&{{a_2}}&{{a_3}}\\ {{b_1}}&{{b_2}}&{{b_3}} \end{array}} \right)$

The determinant of this matrix is:

$\left| {\begin{array}{*{20}{c}} i&j&k\\ {{a_1}}&{{a_2}}&{{a_3}}\\ {{b_1}}&{{b_2}}&{{b_3}} \end{array}} \right| = i\left| {\begin{array}{*{20}{c}} {{a_2}}&{{a_3}}\\ {{b_2}}&{{b_3}} \end{array}} \right| - j\left| {\begin{array}{*{20}{c}} {{a_1}}&{{a_3}}\\ {{b_1}}&{{b_3}} \end{array}} \right| + k\left| {\begin{array}{*{20}{c}} {{a_1}}&{{a_2}}\\ {{b_1}}&{{b_2}} \end{array}} \right| = i\left( {{a_2}{b_3} - {a_3}{b_2}} \right) - j\left( {{a_3}{b_1} - {a_1}{b_3}} \right) + k\left( {{a_1}{b_2} - {a_2}{b_1}} \right)$

where i, j, k, are unit vectors in the direction of the axes x, y, z.

Use of Equation in different problems

What follows are formulae and strategies you may require for question on vectors.

Distance of Point from Plane

The distance between the point A (${a_1},\,{a_2},\,{a_3}$) and the plane defined by  $r \cdot n = d$ is given by:

$l = \frac{{\left| {d - b \cdot n} \right|}}{{\left| n \right|}}$

As a special case, the distance from the origin to a plane is:

${l_O} = \frac{{\left| d \right|}}{{\left| n \right|}}$

The sign of d tells us whether the plane lies above or below the origin.

Distance of a point from a line

There are many ways one may be work out the shortest distance between a point and a line. You must already be aware of at least one from your work in the Mathematics Course. What follows is a method that involves the vector product.

Consider the following case. The line is such that it pass through A, and has the direction vector v, i.e. $r = A + \lambda V$. The point in consideration is S. If we consider the vectors AS, and V, then:

$\begin{array}{*{20}{l}} {\overrightarrow {AS} \times \overrightarrow V = \left| {AS} \right|\left| V \right|\sin \theta \hat n}\\ {\left| {AS \times V} \right| = \left| {\left| {AS} \right|\left| V \right|\sin \theta \hat n} \right|}\\ {\left| {AS \times V} \right| = \left| {AS} \right|\left| V \right|\sin \theta }\\ {\frac{{\left| {AS \times V} \right|}}{{\left| V \right|}} = \left| {AS} \right|\sin \theta }\\ {\frac{{\left| {AS \times V} \right|}}{{\left| V \right|}} = l} \end{array}$

For the above method we realise that the modulus of a unit vector is 1 and that $\left| {AS} \right|\sin \theta $ is the length that is to be found.

Determining whether a line lies in a plane

If we have a line, $ r = \vec a + t\vec b$, and a plane, $ax + by + cz = d$, the method to determine whether the line lies on the plane is given below.

Ex.

Given that:

$\begin{array}{l} r = \left( {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0\\ 1\\ 1 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 1\\ { - 1}\\ 0 \end{array}} \right)\\ x + y + z = 2 \end{array}$

You can either substitute the co-ordinates of a general point directly into the plane, and hence prove that any general point on the plane also satisfies the plane equation:

$\begin{array}{l} \left( {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0\\ 1\\ 1 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 1\\ { - 1}\\ 0 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {0 + t}\\ {1 - t}\\ 1 \end{array}} \right)\\ x + y + z = 2\\ \left( {0 + t} \right) + \left( {1 - t} \right) + 1 = 2\\ 2 = 2 \end{array}$

Alternatively you could prove that some point on the line lies on the plane as well, and that the direction vector is parallel to the plane:

$\begin{array}{l} r = \left( {\begin{array}{*{20}{c}} 0\\ 1\\ 1 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 1\\ { - 1}\\ 0 \end{array}} \right)\\ x + y + z = 2\\ \left( 0 \right) + \left( 1 \right) + 1 = 2\\ 2 = 2\\ n = \left( {\begin{array}{*{20}{c}} 1\\ 1\\ 1 \end{array}} \right);\;\vec b = \left( {\begin{array}{*{20}{c}} 1\\ { - 1}\\ 0 \end{array}} \right)\\ n \cdot b = \left( {\begin{array}{*{20}{c}} 1\\ 1\\ 1 \end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}} 1\\ { - 1}\\ 0 \end{array}} \right) = 1 - 1 + 0 = 0 \end{array}$

The dot product between the normal to the plane and the direction vector of the line is taken to prove that the direction vector, and in turn the line, is parallel to the plane. As is clear, this all follows from what you have already studied and I am teaching nothing new. Following this are also strategies and formulae.

Determine whether the line is parallel or perpendicular to the plane 

To determine whether it is parallel you follow the second method outlined in the previous section, though you will find that the point may not lie in the plane. You could also take a vector lying in the plane, and find the cross product of that vector and the direction vector of the line. If the resultant vector is a multiple of the normal of the plane, the line is parallel to the plane.

To prove that a line is perpendicular to a plane, one finds a vector lying in the plane and take its dot product with the direction vector of the line. If the dot product is zero, the line is parallel.

Angle between line and plane

Let there be a line L and a plane $\pi$:

$\begin{array}{l} L = a + t\vec b\\ \pi :r \cdot n = d \end{array}$

where the second line must be taken to mean, that the plane is defined by the following equation. Also note that the choice of $\pi$ is purely arbitrary and it has nothing to do with the number defined as the ratio of a circle's circumference to its diameter. Regardless, let us say that $\theta$ is the acute angle between the line and the plane. Then:

$\theta  = {\sin ^{ - 1}}\left( {\frac{{n \cdot d}}{{\left| n \right|\left| d \right|}}} \right)$

Angle between two planes

The angle between two planes, is the angle between the two normals to the planes. The angle is therefore:

$\theta  = {\cos ^{ - 1}}\left( {\left| {\frac{{{n_1} \cdot {n_2}}}{{\left| {{n_1}} \right|\left| {{n_2}} \right|}}} \right|} \right)$ 

where ${n_i}$ are the normals to the planes

Line of Intersection of two planes

The determination of the lines of intersection is illustrated through the following example:

Ex. Determine the line of intersection of the planes

$\begin{array}{l} r \cdot \left( {i + j + k} \right) = 2\\ r \cdot \left( {i + 2j + 3k} \right) = 3 \end{array}$

To begin with we write down the cartesian forms:

$\begin{array}{l} x + y + z = 2\\ x + 2y + 3z = 3 \end{array}$

We let $x = \lambda$ ( or you could let y or z be any variable of your choosing )

$\begin{array}{l} y + z = 2 - \lambda \\ 2y + 3z = 3 - \lambda \end{array}$
$\begin{array}{l} 2y + 3z - 2\left( {y + z} \right) = 3 - \lambda - 4 + 2\lambda \Rightarrow z = \lambda - 1\\ 2y + 3z - 3\left( {y + z} \right) = 3 - \lambda - 6 + 3\lambda \Rightarrow y = 3 - 2\lambda \end{array}$

Using this information one can produce the following result:

$\left( {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right) = \left( {\begin{array}{*{20}{c}} \lambda \\ { - 1 + \lambda }\\ {3 - 2\lambda } \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0\\ { - 1}\\ 3 \end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}} 1\\ 1\\ { - 2} \end{array}} \right)$

Shortest distance between two skew lines

Given two lines:

$\begin{array}{l} {r_1} = {a_1} + \lambda {b_1}\\ {r_2} = {a_2} + \lambda {b_2} \end{array}$

The shortest distance is given by:

$l = \frac{{\left| {\left( {{a_1} - {a_2}} \right) \cdot {b_1} \times {b_2}} \right|}}{{\left| {{b_1} \times {b_2}} \right|}}$

The expression in the numerator is called a scalar triple product. You are to compute the vector product first and the scalar product second, but this much should be obvious as it wouldn't work the other way around.

Finding an equation for the common perpendicular to two skew lines

Allow me to explain with an example:

Ex. Find the points of intersection of a line that is mutually perpendicular to the skew lines $L_1$ and $L_2$, and hence find the equation this line.

$\begin{array}{l} {L_1} = \left( {\begin{array}{*{20}{c}} 3\\ { - 1}\\ 2 \end{array}} \right) + s\left( {\begin{array}{*{20}{c}} 1\\ 1\\ 0 \end{array}} \right)\\ {L_2} = \left( {\begin{array}{*{20}{c}} 0\\ 5\\ 2 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 1\\ { - 2}\\ 1 \end{array}} \right) \end{array}$

Sol.

The Line mutually perpendicular to both lines, has its direction vector as the cross product of the direction vectors of $L_1$ and $L_2$. Hence:

$\left( {\begin{array}{*{20}{c}} 1\\ 1\\ 0 \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} 1\\ { - 2}\\ 1 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1\\ { - 1}\\ { - 3} \end{array}} \right)$

General Points of the two lines are:

$\begin{array}{l} {r_1} = \left( {\begin{array}{*{20}{c}} {3 + s}\\ {s - 1}\\ 2 \end{array}} \right)\\ {r_2} = \left( {\begin{array}{*{20}{c}} t\\ {5 - 2t}\\ {2 + t} \end{array}} \right) \end{array}$

The vector that joins these two lines is then:

$\vec v = \left( {\begin{array}{*{20}{c}} {3 + s - t}\\ {s + 2t - 6}\\ { - t} \end{array}} \right)$

This vector should be now perpendicular to the direction vectors of the initial lines.

$\left( {\begin{array}{*{20}{c}} {3 + s - t}\\ {s + 2t - 6}\\ { - t} \end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}} 1\\ 1\\ 0 \end{array}} \right) = 2s + t - 3 = 0$
$\left( {\begin{array}{*{20}{c}} {3 + s - t}\\ {s + 2t - 6}\\ { - t} \end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}} 1\\ { - 2}\\ 1 \end{array}} \right) = - s - 6t + 15 = 0$

Solving these two equations simultaneously, we get:

$s = 3\quad t =  - 3$

Putting these into the general point formula above will give us the points of intersection. We have already found the direction vector for the perpendicular through the cross product. The following procedure to work out the equation need not be showed and thus is left to the reader.