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:

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

This is calculated as follows:

$$ \begin{eqnarray} && \!\!\!\!\!{\underline{i}}\,\begin{vmatrix} \phantom{-}3 & \ 2\,\cr -2 & \ 1\, \end{vmatrix} -{\underline{j}}\,\begin{vmatrix} -1 & 2\,\cr \phantom{-}2 & 1\, \end{vmatrix}+{\underline{k}}\,\begin{vmatrix} -1 &\!\!\phantom{-}3\,\cr \phantom{-}2 &\!\!-2\, \end{vmatrix} \\[10pt] &=& {\underline{i}}\,(3+4)-{\underline{j}}\,(-1-4)+{\underline{k}}\,(2-6) \\[6pt] &=& 7{\underline{i}}+5{\underline{j}}-4{\underline{k}} \\[6pt] \end{eqnarray} $$

The area equals the magnitude of this vector:

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

Detailed breakdown by ChatGPT 

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:

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

This is calculated as follows:

$$ \begin{eqnarray} && \!\!\!\!\!1\begin{vmatrix} \,0 & 1\,\cr \,1 & 5\, \end{vmatrix} -4\begin{vmatrix} \,\phantom{-}3 & 1\,\cr \,-2 & 5\, \end{vmatrix}-2\begin{vmatrix} \,3 &\!\!\!\!\phantom{-}0\,\cr \,-2 &\!\!\!\!\phantom{-}1\, \end{vmatrix} \\[10pt] &=& 1(0-1)-4(15+2)-2(3-0) \\[6pt] &=& -\!1-68-6 \\[6pt] &=& -\!75 \\[6pt] \end{eqnarray} $$

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

Detailed breakdown by ChatGPT 

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:

$$ \begin{gather} \underline{a}\,.\,\underline{b}=\vert\,\underline{a}\,\vert\,\vert\,\underline{b}\,\vert\,\text{cos}\,\theta \\[6pt] 1=\sqrt{6}\,\sqrt{6}\,\text{cos}\,\theta \\[6pt] 1=6\,\text{cos}\,\theta \\[6pt] \text{cos}\,\theta=\small\frac{1}{6}\normalsize \\[6pt] \theta\approx 1.40\:\small\textsf{rad}\normalsize\ \small\textsf{or }\,\normalsize80.4^\circ \\[6pt] \end{gather} $$

Detailed breakdown by ChatGPT 

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

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

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

Writing this in component form will let us obtain 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:

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

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:

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

Detailed breakdown by ChatGPT 

The Cartesian form of the equation of a plane has the same coefficients as the a normal to the plane. To find a normal to the plane, we need the vector product of any two vectors in the plane.

Let's find \(\overrightarrow{\textsf{PQ}}\) and \(\overrightarrow{\textsf{PR}}\):

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

Now we can find the cross product:

$$ \begin{eqnarray} && \!\!\!\!\!\overrightarrow{\textsf{PQ}} \times \overrightarrow{\textsf{PR}} \\[6pt] &=& \begin{vmatrix} \,\phantom{-}\underline{i} & \phantom{-}\underline{j} & \phantom{-}\underline{k}\phantom{.}\cr \,-5 & -3 & \phantom{-}6\phantom{.}\cr \,-1 & -1 & \phantom{-}4\phantom{.} \end{vmatrix} \\[10pt] &=& \underline{i}\,\begin{vmatrix} \,-3 & \ 6\,\cr \,-1 & \ 4\, \end{vmatrix} -\underline{j}\,\begin{vmatrix} \,-5 & 6\,\cr \,-1 & 4\, \end{vmatrix}+\underline{k}\,\begin{vmatrix} \,-5 &\ \!\!\!\!\!-\!3\,\cr \,-1 &\ \!\!\!\!\!-\!1\, \end{vmatrix} \\[10pt] &=& \underline{i}\,(-12+6)-\underline{j}\,(-20+6)+\underline{k}\,(5-3) \\[6pt] &=& -\!6\underline{i}+14\underline{j}+2\underline{k} \\[6pt] \end{eqnarray} $$

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\(\,(3,\,0,\,-1)\) looks easiest, so \(\raise 0.2pt{c=3(3)\!-\!(-1)=10}\) and the equation of the plane is \(\raise 0.2pt{3x-7y-z=10\small.}\)

Detailed breakdown by ChatGPT 

There are several methods to find the equation of the line of intersection of two planes.

Every method is just a different way to solve a system of two linear equations with three variables.

Our preferred method is to create the augmented matrix and use elementary row operations to put it into reduced row echelon form, which means that the bottom-left entry is \(0\) and the first non-zero entry in each row is \(1\small.\)

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

Now we use back substitution to obtain our equation in parametric form.

Start by letting \(z=t\small,\) say. This will be our parameter.

Then, from row 2:

$$ \begin{gather} y-\small\frac34\normalsize z=-2 \\[5pt] y-\small\frac34\normalsize t=-2 \\[5pt] y=-2+\small\frac34\normalsize t \end{gather} $$

Substituting into row 1:

$$ \begin{gather} x-\small\frac12\normalsize y + \small\frac12\normalsize z=2 \\[6pt] x-\small\frac12\normalsize (-2+\small\frac34\normalsize t)+\small\frac12\normalsize t=2 \\[6pt] x+1-\small\frac38\normalsize t+\small\frac12\normalsize t=2 \\[6pt] x=1-\small\frac18\normalsize t \end{gather} $$

So parametric equations of the line of intersection are:

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

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

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

Detailed breakdown by ChatGPT 

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.

$$ \small\frac{x-3}{4}\normalsize = \small\frac{y-2}{-1}\normalsize = \small\frac{z+1}{2}\normalsize =t\small,\,\small\textsf{say.} $$

We rearrange these to obtain expressions for \(x\small,\) \(y\) and \(z\small.\)

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

Now that we have the parametric form of the equation of the line, we can substitute into the Cartesian form of the equation of the plane:

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

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 \(\left(-1,\,3,\,-3\right)\small.\)

Detailed breakdown by ChatGPT 

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:

$$ \begin{eqnarray} && \!\!\!\!\!5\begin{vmatrix} \ 1 & \,3\,\cr \ 4 & -1\, \end{vmatrix} -13\begin{vmatrix} \ 2 & \ 3\,\cr \ 1 & -1\, \end{vmatrix}+0\begin{vmatrix} \ 2 &\!\!\!\!\phantom{-}1\,\cr \ 1 &\!\!\!\!\phantom{-}4\, \end{vmatrix} \\[10pt] &=& 5(-1-12)-13(-2-3)+0 \\[6pt] &=& 5(-13)-13(-5) \\[6pt] &=& -\!65+65 \\[6pt] &=& 0 \\[6pt] \end{eqnarray} $$

The geometric interpretation may be expressed in quite a few different ways. The SQA's marking instructions listed 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.

Detailed breakdown by ChatGPT 

(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:

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

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:

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

Detailed breakdown by ChatGPT 

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

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

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. This will be our parameter.

Then, from row 2:

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

Substituting into row 1:

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

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:

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

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

(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: \(20\neq -3\times 1\small.\)

Detailed breakdown by ChatGPT 

(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 \(\underline{d}_1\) 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:

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

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 \(②\) 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 value into \(③\) gives us:

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

So, \(\mu=-6\small,\normalsize\ \lambda=-20\) is the solution for equations \(②\) and \(③\small.\)

However, we need to check whether or not this solution also satisfies 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.

Therefore, the lines \(L_1\) and \(L_2\) do not intersect.

Detailed breakdown by ChatGPT 

(a)  A direction vector for the line \(L\) is \(\underline{d}=2\underline{i}+\underline{j}-\underline{k}\small.\)

The plane is perpendicular to \(L\) so this vector is normal to the plane, and equation of the plane is of the form:

$$ 2x+y-z=d $$

Substituting \(\text{P}(1,1,0)\):

$$ \begin{gather} 2(1)+1-0=c \\[6pt] c=3 \end{gather} $$

The equation of the plane is \(2x+y-z=3\small.\)

(b)  To find the point of intersection \(\text{Q}\) between the line \(L\) and the plane, write the symmetric equation of \(L\) in parametric form:

$$ x=2t-1,\quad y=t+2,\quad z=-t $$

Substitute these expressions into the plane equation:

$$ \begin{gather} 2(2t-1)+(t+2)-(-t)=3 \\[6pt] 4t-2+t+2+t=3 \\[6pt] 6t=3 \\[6pt] t=\frac{1}{2} \end{gather} $$

Substituting \(t=\displaystyle\small\frac{1}{2}\) gives the point of intersection \(\text{Q}\left(0, \displaystyle\small\frac{5}{2}\normalsize, -\displaystyle\small\frac{1}{2}\normalsize\right)\small.\)

The shortest distance is the length of the line segment \(\text{PQ}\):

$$ \begin{eqnarray} \text{PQ} &=& \sqrt{(0-1)^2 + \left(\frac{5}{2}-1\right)^2 + \left(-\frac{1}{2}-0\right)^2} \\[6pt] &=& \sqrt{(-1)^2 + \left(\frac{3}{2}\right)^2 + \left(-\frac{1}{2}\right)^2} \\[6pt] &=& \sqrt{1 + \frac{9}{4} + \frac{1}{4}} \\[6pt] &=& \sqrt{\frac{14}{4}} \\[6pt] &=& \sqrt{\frac{7}{2}} \end{eqnarray} $$

This is the shortest distance because \(\text{PQ}\) is perpendicular to \(L\small.\)

Detailed breakdown by ChatGPT 

(a)  Find vectors lying in \(\pi_1\):

$$ \overrightarrow{\text{AB}} = \underline{b} - \underline{a} = \begin{pmatrix} 6 \\ -5 \\ 4 \end{pmatrix} - \begin{pmatrix} 4 \\ 0 \\ 8 \end{pmatrix} = \begin{pmatrix} 2 \\ -5 \\ -4 \end{pmatrix} $$

$$ \overrightarrow{\text{AC}} = \underline{c} - \underline{a} = \begin{pmatrix} 3 \\ 4 \\ 11 \end{pmatrix} - \begin{pmatrix} 4 \\ 0 \\ 8 \end{pmatrix} = \begin{pmatrix} -1 \\ 4 \\ 3 \end{pmatrix} $$

A normal vector \(\underline{n}\) to \(\pi_1\) is given by \(\overrightarrow{\text{AB}} \times \overrightarrow{\text{AC}}\):

$$ \begin{vmatrix} \,\phantom{-}{\underline{i}} & \phantom{-}{\underline{j}} & \phantom{-}{\underline{k}}\phantom{.}\cr \,\phantom{-}2 & -5 & -4\phantom{.}\cr \,-1 & \phantom{-}4 & \phantom{-}3\phantom{.} \end{vmatrix} $$

$$ \begin{eqnarray} \underline{n} &=& \underline{i}\,\begin{vmatrix} -5 & -4\,\cr 4 & 3\, \end{vmatrix} - \underline{j}\,\begin{vmatrix} 2 & -4\,\cr -1 & 3\, \end{vmatrix} + \underline{k}\,\begin{vmatrix} 2 & -5\,\cr -1 & 4\, \end{vmatrix} \\[10pt] &=& \underline{i}\,(-15 - (-16)) - \underline{j}\,(6 - 4) + \underline{k}\,(8 - 5) \\[6pt] &=& \underline{i} - 2\underline{j} + 3\underline{k} \end{eqnarray} $$

The equation of the plane is of the form:

$$ x - 2y + 3z = c $$

Substituting \(\text{A}(4, 0, 8)\):

$$ \begin{gather} 4 - 2(0) + 3(8) = c \\[6pt] c = 28 \end{gather} $$

The equation of \(\pi_1\) is \(x - 2y + 3z = 28\small.\)

(b)  \(\pi_2\) is parallel to \(\pi_1\) and passes through the origin, \(d = 0\small.\)

So the equation of \(\pi_2\) is \(x - 2y + 3z = 0\small.\)

(c) (i)  \(\text{AQ}\) is normal to both planes, so its direction vector is \(\underline{n} = \underline{i} - 2\underline{j} + 3\underline{k}\small.\)

Using \(\text{A}(4, 0, 8)\small,\) the parametric equations are:

$$ x = 4 + t, \quad y = -2t, \quad z = 8 + 3t $$

(c) (ii)  Substitute these into the equation for \(\pi_2\):

$$ \begin{gather} (4 + t) - 2(-2t) + 3(8 + 3t) = 0 \\[6pt] 4 + t + 4t + 24 + 9t = 0 \\[6pt] 14t + 28 = 0 \\[6pt] t = -2 \end{gather} $$

Substituting \(t = -2\) gives:

$$ \begin{eqnarray} x &=& 4 + (-2) = 2 \\[6pt] y &=& -2(-2) = 4 \\[6pt] z &=& 8 + 3(-2) = 2 \end{eqnarray} $$

The coordinates of \(\text{Q}\) are \((2, 4, 2)\small.\)

Detailed breakdown by ChatGPT 

(a) (i)  Find \(\overrightarrow{\text{AB}}\) and \(\overrightarrow{\text{AC}}\):

$$ \overrightarrow{\text{AB}} = \underline{b} - \underline{a} = \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} - \begin{pmatrix} 2 \\ -1 \\ 8 \end{pmatrix} = \begin{pmatrix} -1 \\ 2 \\ -9 \end{pmatrix} $$

$$ \overrightarrow{\text{AC}} = \underline{c} - \underline{a} = \begin{pmatrix} 4 \\ -2 \\ 11 \end{pmatrix} - \begin{pmatrix} 2 \\ -1 \\ 8 \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} $$

(a) (ii)  A normal vector \(\underline{n}\) to the plane is given by \(\overrightarrow{\text{AB}} \times \overrightarrow{\text{AC}}\):

$$ \begin{vmatrix} \,\phantom{-}{\underline{i}} & \phantom{-}{\underline{j}} & \phantom{-}{\underline{k}}\phantom{.}\cr \,-1 & \phantom{-}2 & -9\phantom{.}\cr \,\phantom{-}2 & -1 & \phantom{-}3\phantom{.} \end{vmatrix} $$

$$ \begin{eqnarray} \underline{n} &=& \underline{i}\,\begin{vmatrix} 2 & -9\,\cr -1 & 3\, \end{vmatrix} - \underline{j}\,\begin{vmatrix} -1 & -9\,\cr 2 & 3\, \end{vmatrix} + \underline{k}\,\begin{vmatrix} -1 & 2\,\cr 2 & -1\, \end{vmatrix} \\[10pt] &=& \underline{i}\,(6 - 9) - \underline{j}\,(-3 - (-18)) + \underline{k}\,(1 - 4) \\[6pt] &=& -3\underline{i} - 15\underline{j} - 3\underline{k} \end{eqnarray} $$

Scaling by \(-\frac{1}{3}\) gives \(\underline{n} = \underline{i} + 5\underline{j} + \underline{k}\small.\)

The equation of the plane is of the form:

$$ x + 5y + z = c $$

Substituting \(\text{A}(2, -1, 8)\):

$$ \begin{gather} 2 + 5(-1) + 8 = c \\[6pt] c = 5 \end{gather} $$

The equation of the plane is \(x + 5y + z = 5\small.\)

(b)  Write the symmetric equations of the line in parametric form:

$$ x = t + 1, \quad y = -t - 1, \quad z = 4t - 1 $$

Substitute these into the equation of the plane:

$$ \begin{gather} (t + 1) + 5(-t - 1) + (4t - 1) = 5 \\[6pt] t + 1 - 5t - 5 + 4t - 1 = 5 \\[6pt] -5 = 5 \end{gather} $$

The equation is therefore inconsistent, since \(-5 \neq 5,\) so there are no solutions for \(t\small.\)

Therefore the line and the plane do not intersect.

Detailed breakdown by ChatGPT 

(a)  Set up the augmented matrix:

$$ \begin{eqnarray} \begin{array}{r} \small R_1\normalsize\cr \small R_2\normalsize\cr \small R_3\normalsize \end{array} && \left( \begin{array}{ccc:c} 1 & -1 & 4 & -6 \cr 2 & 3 & 1 & 15 \cr 3 & 2 & -2 & 16 \end{array} \right) \\[10pt] \begin{array}{r} \small\textsf{Frozen}\normalsize\cr \small R_2\!-\!2R_1\normalsize\cr \small R_3\!-\!3R_1\normalsize \end{array} && \left( \begin{array}{ccc:c} 1 & -1 & 4 & -6 \cr 0 & 5 & -7 & 27 \cr 0 & 5 & -14 & 34 \end{array} \right) \\[10pt] \begin{array}{r} \small\textsf{Frozen}\normalsize\cr \small\textsf{Frozen}\normalsize\cr \small R_3\!-\!R_2\normalsize \end{array} && \left( \begin{array}{ccc:c} 1 & -1 & 4 & -6 \cr 0 & 5 & -7 & 27 \cr 0 & 0 & -7 & 7 \end{array} \right) \end{eqnarray} $$

From Row 3:

$$ \begin{gather} -7z = 7 \\[6pt] z = -1 \end{gather} $$

From Row 2:

$$ \begin{gather} 5y - 7(-1) = 27 \\[6pt] 5y + 7 = 27 \\[6pt] 5y = 20 \\[6pt] y = 4 \end{gather} $$

From Row 1:

$$ \begin{gather} x - 4 + 4(-1) = -6 \\[6pt] x - 8 = -6 \\[6pt] x = 2 \end{gather} $$

The point \(\text{T}\) is \((2, 4, -1)\small.\)

(b)  Write \(L_1\) in parametric form:

$$ x = 3t - 4, \quad y = 2t + 7, \quad z = t + 4 $$

Substitute into \(\pi_3\):

$$ \begin{gather} 3(3t - 4) + 2(2t + 7) - 2(t + 4) = 16 \\[6pt] 9t - 12 + 4t + 14 - 2t - 8 = 16 \\[6pt] 11t - 6 = 16 \\[6pt] 11t = 22 \\[6pt] t = 2 \end{gather} $$

Substituting \(t = 2\) gives:

$$ \begin{eqnarray} x &=& 3(2) - 4 = 2 \\[6pt] y &=& 2(2) + 7 = 11 \\[6pt] z &=& 2 + 4 = 6 \end{eqnarray} $$

The point \(\text{P}\) is \((2, 11, 6)\small.\)

(c)  Find direction vector \(\overrightarrow{\text{TP}}\):

$$ \overrightarrow{\text{TP}} = \underline{p} - \underline{t} = \begin{pmatrix} 2 \\ 11 \\ 6 \end{pmatrix} - \begin{pmatrix} 2 \\ 4 \\ -1 \end{pmatrix} = \begin{pmatrix} 0 \\ 7 \\ 7 \end{pmatrix} $$

Scaling gives direction vector \(\underline{d} = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} = \underline{j} + \underline{k}\small.\)

Using \(\text{P}(2, 11, 6)\small,\) parametric equations for \(L_2\) are:

$$ x = 2, \quad y = 11 + \lambda, \quad z = 6 + \lambda $$

Detailed breakdown by ChatGPT 

(a)  Find vectors lying in \(\pi\):

$$ \overrightarrow{\text{PQ}} = \underline{q} - \underline{p} = \begin{pmatrix} 3 \\ 5 \\ 1 \end{pmatrix} - \begin{pmatrix} 2 \\ 3 \\ -4 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 5 \end{pmatrix} $$

$$ \overrightarrow{\text{PR}} = \underline{r} - \underline{p} = \begin{pmatrix} 6 \\ 0 \\ -6 \end{pmatrix} - \begin{pmatrix} 2 \\ 3 \\ -4 \end{pmatrix} = \begin{pmatrix} 4 \\ -3 \\ -2 \end{pmatrix} $$

A normal vector \(\underline{n}\) to \(\pi\) is given by \(\overrightarrow{\text{PQ}} \times \overrightarrow{\text{PR}}\):

$$ \begin{vmatrix} \,\phantom{-}{\underline{i}} & \phantom{-}{\underline{j}} & \phantom{-}{\underline{k}}\phantom{.}\cr \,\phantom{-}1 & \phantom{-}2 & \phantom{-}5\phantom{.}\cr \,\phantom{-}4 & -3 & -2\phantom{.} \end{vmatrix} $$

$$ \begin{eqnarray} \underline{n} &=& \underline{i}\,\begin{vmatrix} 2 & 5\,\cr -3 & -2\, \end{vmatrix} - \underline{j}\,\begin{vmatrix} 1 & 5\,\cr 4 & -2\, \end{vmatrix} + \underline{k}\,\begin{vmatrix} 1 & 2\,\cr 4 & -3\, \end{vmatrix} \\[10pt] &=& \underline{i}\,(-4 - (-15)) - \underline{j}\,(-2 - 20) + \underline{k}\,(-3 - 8) \\[6pt] &=& 11\underline{i} + 22\underline{j} - 11\underline{k} \end{eqnarray} $$

Scaling by \(\frac{1}{11}\) gives \(\underline{n} = \underline{i} + 2\underline{j} - \underline{k}\small.\)

The equation of the plane is of the form:

$$ x + 2y - z = c $$

Substituting \(\text{P}(2, 3, -4)\):

$$ \begin{gather} 2 + 2(3) - (-4) = c \\[6pt] c = 12 \end{gather} $$

The equation of \(\pi\) is \(x + 2y - z = 12\small.\)

(b)  Write \(L\) in parametric form:

$$ x = 2t + 8, \quad y = -t - 2, \quad z = 3t - 2 $$

Substitute into \(\pi\):

$$ \begin{gather} (2t + 8) + 2(-t - 2) - (3t - 2) = 12 \\[6pt] 2t + 8 - 2t - 4 - 3t + 2 = 12 \\[6pt] -3t + 6 = 12 \\[6pt] -3t = 6 \\[6pt] t = -2 \end{gather} $$

Substituting \(t = -2\) gives:

$$ \begin{eqnarray} x &=& 2(-2) + 8 = 4 \\[6pt] y &=& -(-2) - 2 = 0 \\[6pt] z &=& 3(-2) - 2 = -8 \end{eqnarray} $$

The coordinates of \(\text{S}\) are \((4, 0, -8)\small.\)

(c)  Identify the direction vector \(\underline{d}\) of \(L\) and normal vector \(\underline{n}\) of \(\pi\):

$$ \underline{d} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}, \quad \underline{n} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} $$

Calculate the scalar product and magnitudes:

$$ \underline{d}.\underline{n} = (2)(1) + (-1)(2) + (3)(-1) = -3 $$

$$ |\underline{d}| = \sqrt{2^2 + (-1)^2 + 3^2} = \sqrt{14} $$

$$ |\underline{n}| = \sqrt{1^2 + 2^2 + (-1)^2} = \sqrt{6} $$

Let \(\theta\) be the acute angle between \(\underline{d}\) and \(\underline{n}\):

$$ \begin{gather} \cos\theta = \frac{\underline{d}.\underline{n}}{\underline{d}||\underline{n}} = \frac{-3}{\sqrt{14}\sqrt{6}} \\[6pt] \theta = \cos^{-1}\left(\frac{3}{\sqrt{84}}\right) \approx 70.9^\circ \end{gather} $$

Finally, the angle between \(L\) and \(\pi\) is \(90^\circ-70.9^\circ=19.1^\circ\small.\)

Detailed breakdown by ChatGPT