Advanced Higher Maths
Vectors

The area of the parallelogram is given by the magnitude of the vector product \(\boldsymbol{\underline{a}}\!\small\times\normalsize\!\boldsymbol{\underline{b}}\)

The vector (cross) product equals the determinant:

$$ \begin{vmatrix} \phantom{-}{\underline{i}} & \phantom{-}{\underline{j}} & \phantom{-}{\underline{k}}\cr -1 & \phantom{-}3 & \phantom{-}2\cr \phantom{-}2 & -2 & \phantom{-}1 \end{vmatrix} $$

This is calculated as follows:

$$ \begin{eqnarray} && {\underline{i}}\,\begin{vmatrix} \phantom{-}3 & \ 2\,\cr -2 & \ 1\, \end{vmatrix} -{\underline{j}}\,\begin{vmatrix} -1 & 2\,\cr \phantom{-}2 & 1\, \end{vmatrix}+{\underline{k}}\,\begin{vmatrix} -1 &\!\!\!\!\phantom{-}3\,\cr \phantom{-}2 &\!\!\!\!-\!2\, \end{vmatrix} \\[6pt] &=& {\underline{i}}(3+4)-{\underline{j}}(-1-4)+{\underline{k}}(2-6) \\[6pt] &=& 7{\underline{i}}+5{\underline{j}}-4{\underline{k}} \\[6pt] \end{eqnarray} $$

The area equals the magnitude of this vector:

$$ \begin{eqnarray} \small\textsf{Area}&=& \sqrt{7^2+5^2+(-4)^2\:} \\[6pt] &=& \sqrt{49+25+16\:} \\[6pt] &=& \sqrt{90\ } \\[6pt] &=& 3\sqrt{10\ }\:\small\textsf{square units} \\[6pt] \end{eqnarray} $$

The volume of the parallelepiped is given by the magnitude of the scalar triple product \(\boldsymbol{a}.\!(\boldsymbol{b}\!\small\times\normalsize\!\boldsymbol{c})\)

It can be shown that this equals the determinant:

$$ \begin{vmatrix} \phantom{-}1 & \phantom{-}4 & -2\cr \phantom{-}3 & \phantom{-}0 & \phantom{-}1\cr -2 & \phantom{-}1 & \phantom{-}5 \end{vmatrix} $$

This is calculated as follows:

$$ \begin{eqnarray} && 1\,\begin{vmatrix} 0 & 1\,\cr 1 & 5\, \end{vmatrix} -4\,\begin{vmatrix} \phantom{-}3 & 1\,\cr -2 & 5\, \end{vmatrix}-2\,\begin{vmatrix} 3 &\!\!\!\!\phantom{-}0\,\cr -2 &\!\!\!\!\phantom{-}1\, \end{vmatrix} \\[6pt] &=& 1(0-1)-4(15+2)-2(3-0) \\[6pt] &=& -\!1-68-6 \\[6pt] &=& -\!75 \\[6pt] \end{eqnarray} $$

The volume of the parallelepiped is therefore \(\vert\!-\!75\vert = 75\) cubic units.

We have been given the Cartesian forms of the equations of these two planes. It is important to understand that in the Cartesian form, the coefficients of \(x\small,\) \(y\) and \(\raise 0.2pt{z}\) define the normal to the plane. Further, the angle between the normals to two planes equals the angle between the planes themselves.

Once the above is understood, this is barely more than a Higher scalar product question. We are finding the angle between \(\boldsymbol{i}+2\boldsymbol{j}-\boldsymbol{k}\) and \(2\boldsymbol{i}-\boldsymbol{j}-\boldsymbol{k}\small.\) For convenience, let's refer to these as \(\boldsymbol{a}\) and \(\boldsymbol{b}\) respectively.

Now we need the scalar product:

$$ \begin{eqnarray} \underline{a}.\underline{b} &=& 1(2)+2(-1)-1(-1)\\[6pt] &=& 2-2+1\\[6pt] &=& 1 \end{eqnarray} $$

Next, we need the magnitude of each vector:

$$ \begin{eqnarray} \vert \underline{a} \vert &=& \sqrt{1^2+2^2+(-1)^2} \\[6pt] &=& \sqrt{1+4+1} \\[6pt] &=& \sqrt{6} \\[12pt] \vert \underline{b} \vert &=& \sqrt{2^2+(-1)^2+(-1)^2} \\[6pt] &=& \sqrt{4+1+1} \\[6pt] &=& \sqrt{6} \\[12pt] \end{eqnarray} $$

Finally, we use the definition of scalar product to find angle \(\theta\small,\) say:

$$ \begin{eqnarray} \underline{a}.\underline{b} &=& \vert \underline{a} \vert\underline{b} \vert\ cos\,\theta \\[6pt] 1 &=& \sqrt{6}\,\sqrt{6}\,cos\,\theta \\[6pt] 1 &=& 6\,cos\,\theta \\[6pt] cos\,\theta &=& \small\frac{1}{6}\normalsize \\[6pt] \theta &\approx& 1.40\:\small\textsf{(radians, to 3 s.f.)}\normalsize \\[6pt] && \small\textsf{or }\,\normalsize80.4^\circ\:\small\textsf{(degrees, to 1 d.p.)}\normalsize \\[6pt] \end{eqnarray} $$

First we obtain the vector equation, which we then split into the parametric form, and finally we convert that into symmetric form.

$$ \begin{eqnarray} \small\overrightarrow{\textsf{AB}}\normalsize &=& \small\overrightarrow{\textsf{OB}}\normalsize-\small\overrightarrow{\textsf{OA}} \\[6pt] &=& \left( \begin{matrix} -2 \\ \phantom{-}3 \\ \phantom{-}7 \end{matrix}\right) - \left( \begin{matrix} \phantom{-}4 \\ -1 \\ \phantom{-}2 \end{matrix}\right) \\[6pt] &=& \left( \begin{matrix} -6 \\ \phantom{-}4 \\ \phantom{-}5 \end{matrix}\right) \end{eqnarray} $$

For any point P\((\raise 0.2pt{x,y,z})\) on the line, \(\small\overrightarrow{\textsf{OP}}\normalsize=\small\overrightarrow{\textsf{OA}}\normalsize+t\small\,\overrightarrow{\textsf{AB}}\small,\) where \(\raise 0.1pt{t\in\mathbb R}\) is a parameter. This is the vector form of the equation of the line.

Writing this in component form will give us the parametric form:

$$ \left( \begin{matrix} \small\,\normalsize x\small\,\normalsize \\ \small\,\normalsize y\small\,\normalsize \\ \small\,\normalsize z\small\,\normalsize \end{matrix}\right) = \left( \begin{matrix} \phantom{-}4 \\ -1 \\ \phantom{-}2 \end{matrix}\right)+t\left( \begin{matrix} -6 \\ \phantom{-}4 \\ \phantom{-}5 \end{matrix}\right) $$

So the parametric equations are:

$$ \begin{eqnarray} x &=& 4-6t \\[6pt] y &=& -\!1+4t\\[6pt] z &=& 2+5t\\[6pt] \end{eqnarray} $$

The symmetric form is when we express each of these with \(t\) as their subject:

$$ \begin{eqnarray} x=4-6t &\implies& t=\small\frac{x-4}{-6} \\[6pt] y=-1+4t &\implies& t=\small\frac{y+1}{4} \\[6pt] z=2+5t &\implies& t=\small\frac{z-2}{5} \end{eqnarray} $$

Equating these gives us the symmetric form:

$$ \small\frac{x-4}{-6}\normalsize =\small\frac{y+1}{4}\normalsize =\small\frac{z-2}{5} $$

As noted in Example 3 above, the Cartesian form of the equation of a plane has coefficients equal to the components of a normal to the plane. So in order to find a normal to this plane, we need the vector product of any two vectors in the plane.

So let's find \(\small\overrightarrow{\textsf{PQ}}\normalsize\) and \(\small\overrightarrow{\textsf{PR}}\):

$$ \begin{eqnarray} \small\overrightarrow{\textsf{PQ}}\normalsize &=& \small\overrightarrow{\textsf{OQ}}\normalsize-\small\overrightarrow{\textsf{OP}} \\[6pt] &=& \left( \begin{matrix} -1 \\ -2 \\ \phantom{-}1 \end{matrix}\right) - \left( \begin{matrix} \phantom{-}4 \\ \phantom{-}1 \\ -5 \end{matrix}\right) \\[6pt] &=& \left( \begin{matrix} -5 \\ -3 \\ \phantom{-}6 \end{matrix}\right) \\[12pt] \small\overrightarrow{\textsf{PR}}\normalsize &=& \small\overrightarrow{\textsf{OR}}\normalsize-\small\overrightarrow{\textsf{OP}} \\[6pt] &=& \left( \begin{matrix} \phantom{-}3 \\ \phantom{-}0 \\ -1 \end{matrix}\right) - \left( \begin{matrix} \phantom{-}4 \\ \phantom{-}1 \\ -5 \end{matrix}\right) \\[6pt] &=& \left( \begin{matrix} -1 \\ -1 \\ \phantom{-}4 \end{matrix}\right) \\[12pt] \end{eqnarray} $$

Now we can find the cross product:

$$ \begin{eqnarray} && \small\overrightarrow{\textsf{PQ}}\normalsize \times \small\overrightarrow{\textsf{PR}}\normalsize \\[6pt] &=& \begin{vmatrix} \phantom{-}\underline{i} & \phantom{-}\underline{j} & \phantom{-}\underline{k}\cr -5 & -3 & \phantom{-}6\cr -1 & -1 & \phantom{-}4 \end{vmatrix} \\[6pt] &=& \underline{i}\,\begin{vmatrix} -3 & \ 6\,\cr -1 & \ 4\, \end{vmatrix} -\underline{j}\,\begin{vmatrix} -5 & 6\,\cr -1 & 4\, \end{vmatrix}+\underline{k}\,\begin{vmatrix} -5 &\ \!\!\!\!\!-\!3\,\cr -1 &\ \!\!\!\!\!-\!1\, \end{vmatrix} \\[6pt] &=& \underline{i}(-12+6)-\underline{j}(-20+6)+\underline{k}(5-3) \\[6pt] &=& -6\underline{i}+14\underline{j}+2\underline{k} \\[6pt] \end{eqnarray} $$

Any scalar multiple of this vector is a normal to the plane. Because the components are all even numbers, and the leading coefficient is negative, it makes sense to halve and negate it, so we will use the normal vector \(3\underline{i}-7\underline{j}-\underline{k}\small.\)

The Cartesian equation of the plane is therefore \(\raise 0.2pt{3x-7y-z=c\small,}\) where \(\raise 0.2pt{c}\) is a constant that can be found simply by subsituting the \(\raise 0.2pt{x\small,}\) \(\raise 0.2pt{y}\) and \(\raise 0.2pt{z}\) coordinates of any of the points P, Q or R.

R looks easiest, so \(\raise 0.2pt{c=3(3)\!-\!(-1)=10}\) and the equation of the plane is therefore \(\raise 0.2pt{3x-7y-z=10\small.}\)

Our preferred method is to create the augmented matrix and reduce it using elementary row operations into reduced row echelon form (where the first entry in the second row is zero and the first non-zero entry in each row is unity).

$$ \begin{eqnarray} \begin{array}{r} \small R_1\normalsize\cr \small R_2\normalsize \end{array} && \left( \begin{array}{ccc:ccc} 2 & -1 & 1 & 4 \cr 4 & 2 & -1 & 0 \end{array} \right) \\[10pt] \begin{array}{r} \small\textsf{Frozen}\normalsize\cr \small R_2\!-\!2R_1\normalsize \end{array} && \left( \begin{array}{ccc:ccc} 2 & -1 & 1 & 4 \cr 0 & 4 & -3 & -8 \end{array} \right) \\[10pt] \begin{array}{r} \small \frac{1}{2}R_1\normalsize\cr \small \frac{1}{4}R_2\normalsize\cr \end{array} && \left( \begin{array}{ccc:ccc} 1 & -\frac12 & \frac12 & 2 \cr 0 & 1 & -\frac34 & -2 \end{array} \right) \\[10pt] \end{eqnarray} $$

Now we use back substitution to obtain our equation in parametric form. Start by letting \(z=t\small.\)

From row 2:

$$ \begin{gather} y-\small\frac34\normalsize t=-2 \\[4pt] y=-2+\small\frac34\normalsize t \end{gather} $$

From row 1:

$$ \begin{gather} x-\small\frac12\normalsize y + \small\frac12\normalsize z=2 \\[4pt] x-\small\frac12\normalsize (-2+\small\frac34\normalsize t)+\small\frac12\normalsize t=2 \\[4pt] x+1-\small\frac38\normalsize t+\small\frac12\normalsize t=2 \\[4pt] x=1-\small\frac18\normalsize t \end{gather} $$

So parametric equations of the line of intersection are:

$$ x=1-\small\frac18\normalsize t\small,\normalsize\:y=-2+\small\frac34\normalsize t\small,\normalsize\:z=t $$

The fractions can be eliminating by redefining the parameter. Let \(\lambda\!=\!\large\frac{t}{8}\) so \(t\!=\!8\lambda\small.\) Then:

$$ x=1-\lambda\small,\normalsize\:y=-2+6\lambda\small,\normalsize\:z=8\lambda $$

To find the point where a line intersects a plane, we express the equation of the line parametrically and then substitute \(x\small,\) \(y\) and \(z\) into the equation of the plane.

$$ \begin{eqnarray} t=\small\frac{x-3}{4}\normalsize &\implies& x=4t+3 \\[6pt] t=\small\frac{y-2}{-1}\normalsize &\implies& y=-t+2 \\[6pt] t=\small\frac{z+1}{2}\normalsize &\implies& z=2t-1 \\[6pt] \end{eqnarray} $$

Now we substitute:

$$ \begin{gather} 2x+y-z=4\\[6pt] 2(4t+3)+(-t+2)-(2t-1)=4\\[6pt] 8t+6-t+2-2t+1=4\\[6pt] 5t+9=4\\[6pt] 5t=-5\\[6pt] t=-1\\[6pt] \end{gather} $$

So the point of intersection is defined by this value of \(t\) and all that remains is to substitute it back into the parametric equations of the line.

$$ \begin{eqnarray} x &=& 4t+3\\[6pt] &=& 4(-1)+3\\[6pt] &=& -\!4+3\\[6pt] &=& -\!1\\[15pt] y &=& -\!t+2\\[6pt] &=& -(-1)+2\\[6pt] &=& 1+2\\[6pt] &=& 3\\[15pt] z &=& 2t-1\\[6pt] &=& 2(-1) -1\\[6pt] &=& -\!2-1\\[6pt] &=& -\!3\\[15pt] \end{eqnarray} $$

So the point of intersection is \(\left(-1,3,-3\right)\).

The scalar triple product \(\boldsymbol{u}.\!(\boldsymbol{v}\!\small\times\normalsize\!\boldsymbol{w})\) is the determinant:

$$ \begin{vmatrix} \phantom{.}5 & 13 & \phantom{-}0\phantom{.}\cr \phantom{.}2 & \phantom{.}1 & \phantom{-}3\phantom{.}\cr \phantom{.}1 & \phantom{.}4 & -1\phantom{.} \end{vmatrix} $$

This is calculated as follows:

$$ \begin{eqnarray} && 5\,\begin{vmatrix} \ 1 & \,3\,\cr \ 4 & -1\, \end{vmatrix} -13\,\begin{vmatrix} \ 2 & \ 3\,\cr \ 1 & -1\, \end{vmatrix}+0\,\begin{vmatrix} \ 2 &\!\!\!\!\phantom{-}1\,\cr \ 1 &\!\!\!\!\phantom{-}4\, \end{vmatrix} \\[6pt] &=& 5(-1-12)-13(-2-3)+0 \\[6pt] &=& 5(-13)-13(-5) \\[6pt] &=& -\!65+65 \\[6pt] &=& 0 \\[6pt] \end{eqnarray} $$

The geometric interpretation may be expressed in quite a few different ways. The SQA's marking instructions list the following alternatives:

• \(\boldsymbol{u}\) lies in the same plane as the one containing both \(\boldsymbol{v}\) and \(\boldsymbol{w}\small.\)
• \(\boldsymbol{u}\) is parallel to the plane containing \(\boldsymbol{v}\) and \(\boldsymbol{w}\small.\)
• \(\boldsymbol{u}\) is perpendicular to the normal of \(\boldsymbol{v}\) and \(\boldsymbol{w}\small.\)
• All four points O, A, B and C lie in the same plane.
• \(\boldsymbol{u}\) is perpendicular to \(\boldsymbol{v}\times\boldsymbol{w}\small.\)
• The volume of the parallelepiped is zero.
• \(\boldsymbol{u}\small,\) \(\boldsymbol{v}\) and \(\boldsymbol{w}\) are coplanar/linearly dependent.

(a)  \(L_1\) is given in parametric form but \(L_2\) is given in symmetric form. In order to show that the lines intersect, we should convert \(L_2\) into parametric form. We may use anything except \(\lambda\) as the parameter for \(L_2\small.\) Let's use \(\mu\small.\)

$$ \begin{eqnarray} \mu=\small\frac{x-3}{-2}\normalsize &\implies& x=3-2\mu \\[6pt] \mu=\small\frac{y-8}{1}\normalsize &\implies& y=8+\mu \\[6pt] \mu=\small\frac{z+1}{3}\normalsize &\implies& z=-1+3\mu \\[6pt] \end{eqnarray} $$

So if there is a point of intersection:

$$ \begin{eqnarray} 4+3\lambda &=& 3-2\mu \\[6pt] 2+4\lambda &=& 8+\mu \\[6pt] -7\lambda &=& -\!1+3\mu \\[6pt] \end{eqnarray} $$

We choose any two of these equations and solve for \(\lambda\) and \(\mu\small.\) Let's just use the first two. However, we must then verify that these values of \(\lambda\) and \(\mu\) also satisfy the third equation. If they don't, there is no point of intersection.

$$ \begin{eqnarray} 3\lambda+2\mu&=& -\!1 &\:\:\:\:①\\[6pt] 4\lambda-\mu &=& 6 &\:\:\:\:②\\[12pt] 8\lambda-2\mu &=& 12 &\:\:\:\:2\!\times\!②=③\\[12pt] 11\lambda &=& 11 &\:\:\:\:①+③\\[6pt] \lambda &=& 1 &\:\\[6pt] \end{eqnarray} $$

Substituting \(\lambda\!=\!1\) into both (not either) of the two equations gives \(\mu\!=\!-\!2\small.\)

Then we must verify these values in \(③\small.\)

$$ \begin{eqnarray} -7(1)&=&-1+3(-2)\\[6pt] -7&=&-1-6\\[6pt] -7&=&-7\\[6pt] \end{eqnarray} $$

So \(\lambda\!=\!1\small,\) \(\mu\!=\!-\!2\) satisfies all three equations. This proves that there is a point of intersection.

To find the point of intersection, we simply substitute one of the parameters back into the parametric equations of the relevant line. Let's use \(\lambda\small.\)

$$ \begin{eqnarray} x &=&4+3\lambda = 4+3=7\\[6pt] y &=&2+4\lambda = 2+4=6 \\[6pt] z &=&-7\lambda = -7\\[6pt] \end{eqnarray} $$

So the point of intersection is \((7,6,-\!7)\small.\)

(b)  To find the angle between the two lines, we need direction vectors within each line. These are just the coefficients of the parameters.

So for \(L_1\small,\) we have \({\underline{d}_1}=3{\underline{i}}+4{\underline{j}}-7{\underline{k}}\)
and for \(L_2\small,\) we have \({\underline{d}_2}=-2{\underline{i}}+{\underline{j}}+3{\underline{k}}\)

We can use scalar product to find the angle \(\theta\small,\) say, between these vectors:

$$ \begin{eqnarray} \underline{d}_1.\underline{d}_2 &=& 3(-2)+4(1)-7(3)\\[6pt] &=& 2-6+4-21\\[6pt] &=& -\!23 \\[6pt] \end{eqnarray} $$

Next, we need the magnitude of each vector:

$$ \begin{eqnarray} \vert \underline{d}_1 \vert &=& \sqrt{3^2+4^2+(-7)^2} \\[6pt] &=& \sqrt{9+16+49} \\[6pt] &=& \sqrt{74} \\[12pt] \vert \underline{d}_2 \vert &=& \sqrt{(-2)^2+1^2+3^2} \\[6pt] &=& \sqrt{4+1+9} \\[6pt] &=& \sqrt{14} \\[12pt] \end{eqnarray} $$

Finally, we use the definition of scalar product to find angle \(\theta\small,\) say:

$$ \begin{eqnarray} \underline{d}_1.\underline{d}_2 &=& \vert \underline{d}_1 \vert\ \underline{d}_2 \vert\ cos\,\theta \\[6pt] -23 &=& \sqrt{74}\,\sqrt{14}\,cos\,\theta \\[6pt] cos\,\theta &=& \small\frac{-23}{\sqrt{74}\sqrt{14}}\normalsize \\[6pt] \theta &\approx& 135.6^\circ \\[6pt] \end{eqnarray} $$

(a)  We apply EROs to reduce the augmented matrix to row echelon form:

$$ \begin{eqnarray} \begin{array}{r} \small R_1\normalsize\cr \small R_2\normalsize\cr \small R_3\normalsize \end{array} && \left( \begin{array}{ccc:ccc} 1 & -2 & 1 & -4\cr 3 & -5 & -2 & 1\cr -7 & 11 & a & -11 \end{array} \right) \\[10pt] \begin{array}{r} \small\textsf{Frozen}\normalsize\cr \small R_2\!-\!3R_1\normalsize \cr \small R_3\!+7\!R_1\normalsize \end{array} && \left( \begin{array}{ccc:ccc} 1 & -2 & 1 & -4\cr 0 & 1 & -5 & 13\cr 0 & -3 & a+7 & -39 \end{array} \right) \\[10pt] \begin{array}{r} \small\textsf{Frozen}\normalsize\cr \small\textsf{Frozen}\normalsize\cr \small R_3\!+\!3R_2\normalsize \end{array} && \left( \begin{array}{ccc:ccc} 1 & -2 & 1 & -4\cr 0 & 1 & -5 & 13\cr 0 & 0 & a-8 & 0 \end{array} \right) \\[10pt] \end{eqnarray} $$

For the three planes to intersect along a straight line (rather than at a point) we need one degree of redundancy, so the third row must consist of zeros. Therefore \(a\!-\!8\!=\!0\) and so \(a\!=\!8\small.\)

(a)  Let \(z\!=\!t\small,\) say.

From row 2:

$$ \begin{gather} y-5z=13 \\[4pt] y-5t=13 \\[4pt] y=13+5t \end{gather} $$

From row 1:

$$ \begin{gather} x-2y+z=-4 \\[4pt] x-2(13+5t)+t=-4 \\[4pt] x-26-10t+t=-4 \\[4pt] x-9t=-4+26 \\[4pt] x=22+9t \end{gather} $$

So parametric equations of the line of intersection are:

$$ x=22+9t\small,\normalsize\:y=13+5t\small,\normalsize\:z=t $$

(c)  This part is very similar to Example 3, above. We will find the angle between the normals to the planes.

Note that the coefficients of the left hand side of the equation of \(\pi_4\) are all multiples of \(3\) so it is convenient to divide them all by this common factor.

$$ \begin{eqnarray} \underline{n}_1 &=& \underline{i}-2\underline{j}+\underline{k} \\[6pt] \underline{n}_4 &=& -\!3\underline{i}+5\underline{j}+2\underline{k} \end{eqnarray} $$

We need the scalar product:

$$ \begin{eqnarray} \underline{n}_1.\underline{n}_4 &=& 1(-\!3)-2(5)+1(2)\\[6pt] &=& -\!3-10+2\\[6pt] &=& -\!11 \end{eqnarray} $$

Next, we need the magnitude of each vector:

$$ \begin{eqnarray} \vert \underline{n}_1 \vert &=& \sqrt{1^2+(-2)^2+1^2} \\[6pt] &=& \sqrt{1+4+1} \\[6pt] &=& \sqrt{6} \\[12pt] \vert \underline{n}_4 \vert &=& \sqrt{(-3)^2+5^2+2^2} \\[6pt] &=& \sqrt{9+25+4} \\[6pt] &=& \sqrt{38} \\[12pt] \end{eqnarray} $$

Finally, we use the definition of scalar product to find angle \(\theta\small,\) say:

$$ \begin{eqnarray} \underline{n}_1.\underline{n}_4 &=& \vert \underline{n}_1 \vert\ \underline{n}_4 \vert\ cos\,\theta \\[6pt] -11 &=& \sqrt{6}\,\sqrt{38}\,cos\,\theta \\[6pt] \theta &=& cos^{-1}\small\left(\frac{-11}{\sqrt{6}\,\sqrt{38}}\right) \\[6pt] \theta &\approx& 136.8^\circ\:\small\textsf{(to 1 d.p.)}\normalsize \end{eqnarray} $$

However, the question asked us for the acute angle between planes \(\pi_1\) and \(\pi_4\small,\) so this is \(180\!-\!136.8\!=\!43.2^\circ\) (to 1 d.p.)

(d)  Considering normals to \(\pi_2\) and \(\pi_4\):

$$ \begin{eqnarray} \underline{n}_2&=&\: 3\underline{i}-5\underline{j}-2\underline{k} \\[6pt] \underline{n}_4&=&\: -\!3\underline{i}+5\underline{j}+2\underline{k} \end{eqnarray} $$

We can see that \(\underline{n}_4=-\underline{n}_2\) so these planes are parallel.

The planes do not coincide because the right hand sides are not equal: \(20\neq -3\times 1\small.\)

(a)  We are only required to verify this line of intersection, not to obtain it for ourselves. So the procedure is to substitute the parametric equations of the line into both of the equations of the plane, to show that the line lies on both planes.

$$ \begin{eqnarray} \pi_{1}&:&\: 2x-3y-z=9 \\[6pt] &&\: 2(2\lambda+3)-3(\lambda-1)-\lambda=9 \\[6pt] &&\: 4\lambda+6-3\lambda+3-\lambda=9 \\[6pt] &&\: 6+3=9 \\[6pt] &&\: 9=9\ \small\textsf{(true)} \\[12pt] \pi_{2}&:&\: x+y-3z=2 \\[6pt] &&\: (2\lambda+3)+(\lambda-1)-3\lambda=2 \\[6pt] &&\: 2\lambda+3+\lambda-1-3\lambda=2 \\[6pt] &&\: 3-1=2 \\[6pt] &&\: 2=2\ \small\textsf{(true)} \\[12pt] \end{eqnarray} $$

\(L_1\) therefore lies on both \(\pi_{1}\) and \(\pi_{2}\small.\) So \(L_1\) is the line of intersection of the planes \(\pi_{1}\) and \(\pi_{2}\small.\)

(b)  The coefficients of \(\lambda\) in the parametric equations of \(L_1\) give us its direction vector: \(2\underline{i}+\underline{j}+\underline{k}=\underline{d}_1\small,\) say.

The normal to the plane \(\pi_3\) is given by the coefficients of its Cartesian equation. So the normal vector is \(-2\underline{i}+4\underline{j}+3\underline{k}=\underline{d}_2\small,\) say.

We will find the angle between what we have called \(\underline{d}_1\) and \(\underline{d}_2\small,\) but it is important to remember that the actual angle between \(L_1\) and \(\pi_3\) will be the complement of this angle, because the normal is at right angles to the plane.

First we use the component definition of scalar product:

$$ \begin{eqnarray} \underline{d}_1.\underline{d}_2 &=& 2(-2)+1(4)+1(3)\\[6pt] &=& -4+4+3\\[6pt] &=& 3 \\[6pt] \end{eqnarray} $$

Next, we need the magnitude of each vector:

$$ \begin{eqnarray} \vert \underline{d}_1 \vert &=& \sqrt{2^2+1^2+1^2} \\[6pt] &=& \sqrt{4+1+1} \\[6pt] &=& \sqrt{6} \\[12pt] \vert \underline{d}_2 \vert &=& \sqrt{(-2)^2+4^2+3^2} \\[6pt] &=& \sqrt{4+16+9} \\[6pt] &=& \sqrt{29} \\[12pt] \end{eqnarray} $$

Finally, we use the definition of scalar product to find angle \(\theta\small,\) say:

$$ \begin{eqnarray} \underline{d}_1.\underline{d}_2 &=& \vert \underline{d}_1 \vert\ \underline{d}_2 \vert\ cos\,\theta \\[6pt] 3 &=& \sqrt{6}\,\sqrt{29}\,cos\,\theta \\[6pt] cos\,\theta &=& \small\frac{3}{\sqrt{6}\sqrt{29}}\normalsize \\[6pt] \theta &\approx& 76.9^\circ \\[6pt] \end{eqnarray} $$

So the angle between \(L_1\) and \(\pi_3\) is \(90-76.9=13.1^\circ\) (correct to 1 d.p.)

(c)  The direction vector of \(L_2\) is the normal to \(\pi_3\) so it is \(-2\underline{i}+4\underline{j}+3\underline{k}\small.\)

\(L_2\) passes through P\((1,3,-2)\) so its parametric equations are:

$$ \begin{eqnarray} x &=& 1-2\mu \\[6pt] y &=& 3+4\mu\\[6pt] z &=& -\!2+3\mu\\[6pt] \end{eqnarray} $$

So if there is a point of intersection:

$$ \begin{eqnarray} 2\lambda+3 &=& 1-2\mu &\:& ①\\[6pt] \lambda-1 &=& 3+4\mu &\:& ②\\[6pt] \lambda &=& -\!2+3\mu &\:& ③\\[6pt] \end{eqnarray} $$

Considering equations ② and ③ (because they look simplest), equation ② rearranges to:

$$ \lambda = 4+4\mu $$

Substituting this into ③, we obtain:

$$ \begin{eqnarray} 4+4\mu &=& -\!2+3\mu \\[6pt] 4\mu-3\mu &=& -\!2-4 \\[6pt] \mu &=& -\!6\\[6pt] \end{eqnarray} $$

Substituting this into ③ gives us:

$$ \lambda=-\!2+3(-6)=-20 $$

Checking this into ②:

$$ \begin{eqnarray} \lambda-1 &=& 3+4\mu \\[6pt] -20-1 &=& 3+4(-6) \\[6pt] -21 &=& 3-24 \\[6pt] -21 &=& -21 \\[6pt] \end{eqnarray} $$

This is true, so these values of \(\lambda\) and \(\mu\) work for ② and ③.

Now we need to check if these values are consistent with equation ①:

$$ \begin{eqnarray} 2\lambda+3 &=& 1-2\mu \\[6pt] 2(-20)+3 &=& 1-2(-6) \\[6pt] -40+3 &=& 1+12 \\[6pt] -37 &=& 13 \\[6pt] \end{eqnarray} $$

This is false, so there are no values of \(\lambda\) and \(\mu\) that are consistent with all three equations, and hence the lines \(L_1\) and \(L_2\) do not intersect.