Higher Maths
Vectors

(a)  For \(\overrightarrow{\textsf{RT}}\small,\) we should start at R and go via S to T, as we have been given the vectors \(\overrightarrow{\textsf{RS}}\) and \(\overrightarrow{\textsf{ST}}\small.\)

So \(\overrightarrow{\textsf{RT}} = -2\underline i -\underline j +6\underline k.\)

(b)  For \(\overrightarrow{\textsf{RP}}\small,\) we should start at R and go via T to P, because the only vector involving P that we know is \(\overrightarrow{\textsf{PT}}\small.\)

So \(\overrightarrow{\textsf{RP}} = -4\underline i -2\underline j +9\underline k.\)

This question revises 2D vector pathways from N5, which are also needed for Higher, eg. 2015 P2 Q6 .

(a)  Let's go from R to P via Q:

(b)  Our first job is to decide a route from R to S.

Looking back at part (a) should be a hint to just go from R to S via P.

Before we begin, note that the question tells us that \(\overrightarrow{\textsf{RQ}}=2\,\overrightarrow{\textsf{SP}}\small.\) We will need to use that fact in our working:

Note that we can use any two vectors for the following working. You might prefer to use \(\overrightarrow{\textsf{AB}}\) and \(\overrightarrow{\textsf{BC}}\) if that feels more natural to you.

$$ \begin{eqnarray} \overrightarrow{\textsf{AB}} &=& \overrightarrow{\textsf{OB}}-\overrightarrow{\textsf{OA}}\\[6pt] &=& \left( \begin{matrix} \phantom{-}2\, \\ -1\, \\ \phantom{-}4\, \end{matrix}\right) - \left( \begin{matrix} -1\, \\ \phantom{-}3\, \\ \phantom{-}0\, \end{matrix}\right)\\[6pt] &=& \left( \begin{matrix} \phantom{-}3\, \\ -4\, \\ \phantom{-}4\, \end{matrix}\right)\\[12pt] \overrightarrow{\textsf{AC}} &=& \overrightarrow{\textsf{OC}}-\overrightarrow{\textsf{OA}}\\[6pt] &=& \left( \begin{matrix} -7\, \\ \phantom{.}11\, \\ -8\, \end{matrix}\right) - \left( \begin{matrix} -1\, \\ \phantom{-}3\, \\ \phantom{-}0\, \end{matrix}\right)\\[6pt] &=& \left( \begin{matrix} -6\, \\ \phantom{-}8\, \\ -8\, \end{matrix}\right)\\[6pt] &=& -\!2\left( \begin{matrix} \phantom{-}3\, \\ -4\, \\ \phantom{-}4\, \end{matrix}\right) \\[12pt] \end{eqnarray} $$

\(\overrightarrow{\textsf{AC}}=-\small\,\normalsize 2\overrightarrow{\textsf{AB}}\) so \(\overrightarrow{\textsf{AC}}\) and \(\overrightarrow{\textsf{AB}}\) are parallel. A is a common point, so A, B and C are collinear.

Note: The course specification  specifically states that candidates should "include the phrases ‘parallel’ and ‘common point’ in their answers to show collinearity," so please be careful to use this wording.

Like the previous example, we can use any two vectors. You might prefer to use \(\overrightarrow{\textsf{PQ}}\) and \(\overrightarrow{\textsf{QR}}\small\) if that feels more natural to you. The working will be different, but the conclusion will be the same.

$$ \begin{eqnarray} \overrightarrow{\textsf{PQ}} &=& \overrightarrow{\textsf{OQ}}-\overrightarrow{\textsf{OP}}\\[6pt] &=& \left( \begin{matrix} \phantom{-}0\, \\ -5\, \\ \phantom{-}5\, \end{matrix}\right) - \left( \begin{matrix} \phantom{-}2\, \\ -6\, \\ \phantom{-}8\, \end{matrix}\right)\\[6pt] &=& \left( \begin{matrix} -2\, \\ \phantom{-}1\, \\ -3\, \end{matrix}\right)\\[12pt] \overrightarrow{\textsf{PR}} &=& \overrightarrow{\textsf{OR}}-\overrightarrow{\textsf{OP}}\\[6pt] &=& \left( \begin{matrix} \phantom{-}8\, \\ -9\, \\ \phantom{-}0\, \end{matrix}\right) - \left( \begin{matrix} \phantom{-}2\, \\ -6\, \\ \phantom{-}8\, \end{matrix}\right)\\[6pt] &=& \left( \begin{matrix} \phantom{-}6\, \\ -3\, \\ -8\, \end{matrix}\right)\\[12pt] \end{eqnarray} $$

Dividing the components of \(\overrightarrow{\textsf{PQ}}\) and \(\overrightarrow{\textsf{PR}}\) to compare them:

\(\:\:\:\underline i\): \(\:\:\:6\div -2=-3\)
\(\:\:\:\underline j\): \(\:-3\div 1=-3\)
\(\:\:\:\underline k\): \(\:-8\div -3 \neq -3\)

The results of the division are not all equal so \(\overrightarrow{\textsf{PR}}\) is not a multiple of \(\overrightarrow{\textsf{PQ}}\small.\) This tells us that the two vectors are not parallel, and for that reason points P, Q and R cannot be collinear.

(a)  Because G divides FH, it makes sense to choose two vectors that both involve G.

$$ \begin{eqnarray} \overrightarrow{\textsf{FG}} &=& \overrightarrow{\textsf{OG}}-\overrightarrow{\textsf{OF}}\\[6pt] &=& \left( \begin{matrix} \!\phantom{-}0\, \\ \!\phantom{-}1\, \\ \!\phantom{-}0\, \end{matrix}\right) - \left( \begin{matrix} -3\, \\ -5\, \\ \phantom{-}9\, \end{matrix}\right)\\[6pt] &=& \left( \begin{matrix} \phantom{-}3\, \\ \phantom{-}6\, \\ -9\, \end{matrix}\right)\\[12pt] \overrightarrow{\textsf{GH}} &=& \overrightarrow{\textsf{OH}}-\overrightarrow{\textsf{OG}}\\[6pt] &=& \left( \begin{matrix} \phantom{-}2\, \\ \phantom{-}5\, \\ -6\, \end{matrix}\right) - \left( \begin{matrix} \!\phantom{-}0\, \\ \!\phantom{-}1\, \\ \!\phantom{-}0\, \end{matrix}\right)\\[6pt] &=& \left( \begin{matrix} \phantom{-}2\, \\ \phantom{-}4\, \\ -6\, \end{matrix}\right) \\[12pt] \end{eqnarray} $$

From the above, \(\overrightarrow{\textsf{FG}}=\large\frac{3}{2}\normalsize\,\overrightarrow{\textsf{GH}}\small,\) so \(\overrightarrow{\textsf{FG}}\) and \(\overrightarrow{\textsf{GH}}\) are parallel. G is a common point, so F, G and H are collinear.

(b)  With experience, you can usually write down these ratios without any further working, but a more formal method would be to start with the connection between \(\overrightarrow{\textsf{FG}}\) and \(\overrightarrow{\textsf{GH}}\) from part (a) above:

$$ \begin{gather} \overrightarrow{\textsf{FG}}=\frac{3}{2}\,\overrightarrow{\textsf{GH}} \\[8pt] \frac{\overrightarrow{\textsf{FG}}}{\overrightarrow{\textsf{GH}}}=\frac{3}{2} \\[12pt] \overrightarrow{\textsf{FG}}:\overrightarrow{\textsf{GH}}=3:2\\[8pt] \end{gather} $$

(a)  We are told that the three points are collinear, so it is sufficient to consider any component.

Let's look at the \(\underline i\) components:

$$ \begin{eqnarray} \overrightarrow{\textsf{AB}}:\overrightarrow{\textsf{BC}} &=& 1-9:-1-1 \\[6pt] &=& -\!8:-2 \\[6pt] &=& \,4:1 \end{eqnarray} $$

(b)  Because \(t\) is a \(\underline j\) component, we can do the same thing with the \(\underline j\) components:

$$ \begin{eqnarray} \overrightarrow{\textsf{AB}}:\overrightarrow{\textsf{BC}} &=& t-(-3):2-t \\[6pt] &=& t+3:2-t \end{eqnarray} $$

Then we can use the ratio \(4:1\) to make a simple linear equation:

$$ \begin{eqnarray} t+3 &=& 4(2-t) \\[6pt] t+3 &=& 8-4t \\[6pt] t+4t &=& 8-3 \\[6pt] 5t &=& 5 \\[6pt] t &=& 1 \end{eqnarray} $$

There are three main methods to divide a line segment in a given ratio. Use whichever you prefer.

Method 1: Stepping out

First we subtract each of the \(x\)-, \(y\)- and \(z\)-coordinates. This calculates the components of the vector \(\overrightarrow{\textsf{RT}}\small.\)

\(x_\textsf{T}-x_\textsf{R}=-3-7=-10\)
\(y_\textsf{T}-y_\textsf{R}=4-(-1)=5\)
\(z_\textsf{T}-z_\textsf{R}=-7-8=-15\)

Then we use the ratio \(2:3\) to split each of these into 'steps':

\(x\!: -10\) splits as \(-4:-6\small.\)
\(y\!: 5\) splits as \(2:3\small.\)
\(z\!: -15\) splits as \(-6:-9\small.\)

Finally, we add each of the 'steps' (i.e. the first numbers in each ratio) onto the corresponding coordinates of R.

\(x_\textsf{R}-4=7-4=3\)
\(y_\textsf{R}+2=-1+2=1\)
\(z_\textsf{R}-6=8-6=2\)

So S has coordinates \((3,\,1,\,2)\small.\)

Method 2: Basic vector algebra

This method starts by stating the ratio, rearranging the resulting equation and finding \(\boldsymbol s\small,\) the position vector of S.

So the coordinates of S are \((3,\,1,\,2).\)

Method 3: The section formula

We don't like the section formula. It's ugly, it's easy to forget and it discourages logical process. We much prefer one of the previous two methods. However, you may use whichever method suits you best!

Let \(m=2\) and \(n=3.\)

$$ \begin{eqnarray} \underline s &=& \frac{n\underline r + m\underline t}{n+m} \\[6pt] &=& \frac{3\underline r + 2\underline t}{2+3} \\[6pt] &=& \small\frac{1}{5}\normalsize\left(3\underline r + 2\underline t\right) \\[6pt] &=& \small\frac{1}{5}\normalsize\left( \begin{matrix} 3(7)+2(-3) \\ 3(-1)+2(4) \\ 3(8)+2(-7) \end{matrix}\right) \\[6pt] &=& \small\frac{1}{5}\normalsize\left( \begin{matrix} 21-6 \\ -3+8 \\ 24-14 \end{matrix}\right) \\[6pt] &=& \small\frac{1}{5}\normalsize\left( \begin{matrix} 15 \\ 5 \\ 10 \end{matrix}\right) \\[6pt] &=& \left( \begin{matrix} \phantom{.}3\phantom{.} \\ \phantom{.}1\phantom{.} \\ \phantom{.}2\phantom{.} \end{matrix}\right) \end{eqnarray} $$

So the coordinates of S are \((3,\,1,\,2).\)

First we find the magnitude of \(\overrightarrow{\textsf{AB}}\small.\)

$$ \begin{eqnarray} \vert \overrightarrow{\textsf{AB}} \vert &=& \sqrt{(-4)^2+(1+6)^2+(-3-1)^2} \\[6pt] &=& \sqrt{(-4)^2+7^2+(-4)^2} \\[6pt] &=& \sqrt{16+49+16} \\[6pt] &=& \sqrt{81} \\[6pt] &=& 9 \end{eqnarray} $$

\(\vert k\,\overrightarrow{\textsf{AB}}\vert=1\) so \(k=\large\frac{1}{9}\small.\)

Questions to do with the angle between vectors always use the scalar product, as defined at the top of this page.

Because \(\underline u\) and \(\underline v\) are perpendicular, we know that \(\underline u\,.\,\underline v=0\small.\) This is because \(\text{cos}\,90^\circ=0\small.\)

$$ \begin{gather} \underline u\,.\,\underline v=0 \\[6pt] -3(2)+2(5)+2n=0 \\[6pt] -6+10+2n=0 \\[6pt] 4+2n=0 \\[6pt] 2n=-4 \\[6pt] n=-2 \end{gather} $$

We have to find the angle at B, so we should find the two vectors that start at B.

$$ \begin{eqnarray} \overrightarrow{\textsf{BA}} &=& \overrightarrow{\textsf{OA}} - \overrightarrow{\textsf{OB}}\\[6pt] &=& \left( \begin{matrix} -7\, \\ -3\, \\ -6\, \end{matrix}\right) - \left( \begin{matrix} \phantom{-}5\, \\ -2\, \\ \phantom{-}6\, \end{matrix}\right)\\[6pt] &=& \left( \begin{matrix} -12\, \\ \phantom{.}-\!1\, \\ -12\, \end{matrix}\right)\\[12pt] \overrightarrow{\textsf{BC}} &=& \overrightarrow{\textsf{OC}} - \overrightarrow{\textsf{OB}}\\[6pt] &=& \left( \begin{matrix} \phantom{-}7\, \\ \phantom{-}3\, \\ -8\, \end{matrix}\right) - \left( \begin{matrix} \phantom{-}5\, \\ -2\, \\ \phantom{-}6\, \end{matrix}\right)\\[6pt] &=& \left( \begin{matrix} \phantom{-}2\, \\ \phantom{-}5\, \\ -14\, \end{matrix}\right)\\[12pt] \end{eqnarray} $$

Now we need the scalar product:

$$ \begin{eqnarray} \overrightarrow{\textsf{BA}}\,.\,\overrightarrow{\textsf{BC}} &=& -\!12(2)-1(5)-12(-14)\\[6pt] &=& -\!24-5+168\\[6pt] &=& \,139 \end{eqnarray} $$

Next, we need the magnitude of each vector:

$$ \begin{eqnarray} \vert\overrightarrow{\textsf{BA}}\vert &=& \sqrt{(-12)^2+(-1)^2+(-12)^2} \\[6pt] &=& \sqrt{144+1+144} \\[6pt] &=& \sqrt{289} \\[6pt] &=& \,17 \\[12pt] \vert\overrightarrow{\textsf{BC}} \vert &=& \sqrt{2^2+5^2+(-14)^2} \\[6pt] &=& \sqrt{4+25+196} \\[6pt] &=& \sqrt{225} \\[6pt] &=& \,15 \\[12pt] \end{eqnarray} $$

Finally, we use the other definition of scalar product to find angle ABC:

$$ \begin{gather} \overrightarrow{\textsf{BA}}\,.\,\overrightarrow{\textsf{BC}}=\vert\overrightarrow{\textsf{BA}} \vert\ \vert\overrightarrow{\textsf{BC}}\vert\ \text{cos}\,\textsf{ABC} \\[6pt] 139=(17)(15)\,\text{cos}\,\textsf{ABC} \\[6pt] 139=255\,\text{cos}\,\textsf{ABC} \\[6pt] \text{cos}\,\textsf{ABC}=\frac{139}{255} \\[6pt] \end{gather} $$ $$ \begin{eqnarray} \angle\textsf{ABC}&=& \text{cos}^{-1}\ \frac{139}{255}\\[6pt] &\approx& 57.0^\circ\:\small\textsf{(to 1 d.p.)}\normalsize \\[6pt] \end{eqnarray} $$

Because \(\underline u\) and \(\underline v\) are perpendicular, we know that \(\underline u\,.\,\underline v=0.\) This is because \(\text{cos}\,90^\circ=0\small.\)

$$ \begin{gather} \underline u\,.\,\underline v=0 \\[6pt] 8(-3)+2t-1(-6)=0 \\[6pt] -24+2t+6=0 \\[6pt] 2t-18=0 \\[6pt] 2t=18 \\[6pt] t=9 \end{gather} $$

(a)  We calculate the dot product using the components:

$$ \begin{eqnarray} \underline{u}\,.\,\underline{v} &=& -\!1\!\times\!-7 + 4\!\times\!8 + (-3)\!\times\!5\\[6pt] &=& 7+32-15\\[6pt] &=& 24 \end{eqnarray} $$

(b)  Now we use the other definition of scalar product, which first requires us to calculate each magnitude:

$$ \begin{eqnarray} \vert\underline{u}\vert &=& \sqrt{(-1)^2+4^2+(-3)^2} \\[6pt] &=& \sqrt{1+16+9} \\[6pt] &=& \sqrt{26}\\[15pt] \vert\underline{v}\vert &=& \sqrt{(-7)^2+8^2+5^2} \\[6pt] &=& \sqrt{49+64+25} \\[6pt] &=& \sqrt{138} \end{eqnarray} $$

Now we can find the angle. Let's call it \(\theta\small.\)

$$ \begin{gather} \underline{u}\,.\,\underline{v}=\vert\underline{u}\vert\ \vert\underline{v}\vert\ \text{cos}\,\theta \\[8pt] 24=\sqrt{26}\,\sqrt{138}\,\text{cos}\,\theta \\[8pt] \text{cos}\,\theta=\frac{24}{\sqrt{26}\,\sqrt{138}} \\[8pt] \end{gather} $$ $$ \begin{eqnarray} \theta &=& \text{cos}^{-1}\ \frac{24}{\sqrt{26}\sqrt{138}} \\[8pt] &\approx& 66.4^\circ\:\small\textsf{(to 1 d.p.)} \\[8pt] \end{eqnarray} $$

(a) (i)  We calculate the scalar product using the components.

$$ \begin{eqnarray} \underline{u}\,.\,\underline{v} &=& p(2p+16)+(-2)\!\times\!(-3)+4\!\times\!6\\[6pt] &=& 2p^2+16p+6+24\\[6pt] &=& 2p^2+16p+30\\[15pt] \end{eqnarray} $$

(a) (ii)  If \(\underline u\) and \(\underline v\) are perpendicular, then \(\underline u\,.\,\underline v=0\) because \(\text{cos}\,90^\circ=0\small.\) So we solve:

$$ \begin{gather} 2p^2+16p+30=0\\[6pt] p^2+8p+15=0\\[6pt] (p+5)(p+3)=0\\[6pt] p+5=0\:\:\small\textsf{or}\normalsize\:\:p+3=0\\[6pt] p=-5\:\:\small\textsf{or}\normalsize\:\:p=-3\\[15pt] \end{gather} $$

(b)  If two vectors are parallel, then one is a scalar multiple of the other.

Examining the \(\underline j\) and \(\underline k\) components of \(\underline u\) and \(\underline v\small,\) we see that \(-3=\large\frac32\normalsize\!\times\!-2\) and \(6=\large\frac32\normalsize\!\times 4\small.\)

So if \(\underline u\) and \(\underline v\) are parallel, then \(\underline v=\large\frac32\normalsize\,\underline u\small.\)

Now we consider the \(\underline i\) components of each vector:

$$ \begin{gather} 2p+16=\small\frac32\normalsize p\\[6pt] 4p+32=3p\\[6pt] 4p-3p=-32\\[6pt] p=-32\\[6pt] \end{gather} $$

We start by expanding the bracket in the usual way:

$$ \begin{gather} \underline{u}\,.\,(\underline{u}+\underline{v})=21 \\[6pt] \underline{u}\,.\,\underline{u} + \underline{u}\,.\,\underline{v}=21 \\[6pt] \end{gather} $$

Now we need to make use of one of the given definitions of scalar product: \( \boldsymbol a .\boldsymbol b = \vert \boldsymbol a\vert\tiny\,\normalsize\vert\boldsymbol b\vert\small\,\normalsize \text{cos}\,\theta\,\small. \)

Note that the angle between \(\underline{u}\) and itself is \(0^\circ\small,\) and that \(\text{cos}\,0^\circ=1\small.\)

So the rest of the solution is as follows:

$$ \begin{gather} \vert\underline{u}\vert\,\vert\underline{u}\vert\,\text{cos}\,0^\circ + \vert\underline{u}\vert\,\vert\underline{v}\vert\,\text{cos}\,\theta^\circ=21 \\[7pt] 4\times 4\times 1\,\,+\,\,4\times 5\,\text{cos}\,\theta=21 \\[7pt] 16 + 20\,\text{cos}\,\theta=21 \\[7pt] 20\,\text{cos}\,\theta=5 \\[6pt] \text{cos}\,\theta=\small\frac{5}{20}\normalsize=\small\frac{1}{4}\normalsize \\[6pt] \end{gather} $$ $$ \begin{eqnarray} \theta &=& \text{cos}\,^{-1}\left(\small\frac{1}{4}\right)\normalsize \\[6pt] &\approx& 75.5^\circ \end{eqnarray} $$

So the angle between vectors \(\underline{u}\) and \(\underline{v}\) is \(75.5^\circ\) (to 1 d.p. accuracy).

Note: If you prefer working in radians, the angle is approximately \(1.32\) rad.

(a) (i)  First we find \(\overrightarrow{\textsf{AB}}\) in component form, and then we find its magnitude.

$$ \begin{eqnarray} \overrightarrow{\textsf{AB}} &=& \overrightarrow{\textsf{OB}}-\overrightarrow{\textsf{OA}}\\[6pt] &=& \left( \begin{matrix} \phantom{-}7\phantom{.} \\ -4\phantom{.} \\ \phantom{-}1\phantom{.} \end{matrix}\right) - \left( \begin{matrix} \phantom{-}4\phantom{.} \\ \phantom{-}2\phantom{.} \\ -5\phantom{.} \end{matrix}\right)\\[6pt] &=& \left( \begin{matrix} \phantom{-}3\phantom{.} \\ -6\phantom{.} \\ \phantom{-}6\phantom{.} \end{matrix}\right)\\[12pt] \end{eqnarray} $$

$$ \begin{eqnarray} \vert \overrightarrow{\textsf{AB}} \vert &=& \sqrt{3^2+(-6)^2+6^2} \\[6pt] &=& \sqrt{9+36+36} \\[6pt] &=& \sqrt{81} \\[6pt] &=& \ 9 \\[12pt] \end{eqnarray} $$

(a) (ii)  \(\vert\overrightarrow{\textsf{AB}}\vert=9\) and we are told that \(\vert\overrightarrow{\textsf{BC}}\vert=6\small.\)

So B divides AC in the ratio \(9:6=3:2\small.\)

(b)  There are several reasonable ways to do this. Here is one suitable method.

Draw yourself a quick sketch if you don't immediately see why \(\overrightarrow{\textsf{AC}}=\large\frac{5}{3}\normalsize\,\overrightarrow{\textsf{AB}}\small.\)

$$ \begin{eqnarray} \overrightarrow{\textsf{OC}} &=& \overrightarrow{\textsf{OA}} + \overrightarrow{\textsf{AC}} \\[6pt] &=& \overrightarrow{\textsf{OA}}+\frac{5}{3}\,\overrightarrow{\textsf{AB}} \\[6pt] &=& \left( \begin{matrix} \phantom{-}4\phantom{.} \\ \phantom{-}2\phantom{.} \\ -5\phantom{.} \end{matrix}\right) + \frac{5}{3}\left( \begin{matrix} \phantom{-}3\phantom{.} \\ -6\phantom{.} \\ \phantom{-}6\phantom{.} \end{matrix}\right) \\[6pt] &=& \left( \begin{matrix} \phantom{-}4\phantom{.} \\ \phantom{-}2\phantom{.} \\ -5\phantom{.} \end{matrix}\right) + \left( \begin{matrix} \phantom{\ -}5\phantom{.} \\ -10\phantom{.} \\ \phantom{-}10\phantom{.} \end{matrix}\right) \\[6pt] &=& \left( \begin{matrix} \phantom{-}9\phantom{.} \\ -8\phantom{.} \\ \phantom{-}5\phantom{.} \end{matrix}\right) \end{eqnarray} $$

So C has coordinates \((9,\,-8,\,5)\small.\)

(a)   This straightforward question was worth 6 marks: 2 for (a) and 4 for (b).

$$ \begin{eqnarray} \overrightarrow{\textsf{AB}} &=& \overrightarrow{\textsf{OB}} - \overrightarrow{\textsf{OA}}\\[6pt] &=& \left( \begin{matrix} -2\phantom{.} \\ \phantom{-}5\phantom{.} \\ \phantom{-}1\phantom{.} \end{matrix}\right) - \left( \begin{matrix} \phantom{.}3\phantom{.} \\ \phantom{.}1\phantom{.} \\ \phantom{.}8\phantom{.} \end{matrix}\right)\\[6pt] &=& \left( \begin{matrix} -5\phantom{.} \\ \phantom{-}4\phantom{.} \\ -7\phantom{.} \end{matrix}\right)\\[12pt] \overrightarrow{\textsf{AC}} &=& \overrightarrow{\textsf{OC}} - \overrightarrow{\textsf{OA}}\\[6pt] &=& \left( \begin{matrix} \phantom{-}7\phantom{.} \\ -6\phantom{.} \\ \phantom{-}3\phantom{.} \end{matrix}\right) - \left( \begin{matrix} \phantom{.}3\phantom{.} \\ \phantom{.}1\phantom{.} \\ \phantom{.}8\phantom{.} \end{matrix}\right)\\[6pt] &=& \left( \begin{matrix} \phantom{-}4\phantom{.} \\ -7\phantom{.} \\ -5\phantom{.} \end{matrix}\right)\\[12pt] \end{eqnarray} $$

(b)   Both vectors from part (a) start at the point A, so there are no complications here. We can just use the vectors \(\overrightarrow{\textsf{AB}}\) and \(\overrightarrow{\textsf{AC}}\) to find angle BAC.

First, we use the components to calculate the scalar product:

$$ \begin{eqnarray} \overrightarrow{\textsf{AB}}\,. \,\overrightarrow{\textsf{AC}} &=& -\!5(4)+4(-7)-7(-5)\\[6pt] &=& -\!20-28+35\\[6pt] &=& -\!13 \end{eqnarray} $$

Next, we find the magnitude of each vector:

$$ \begin{eqnarray} \vert \overrightarrow{\textsf{AB}} \vert &=& \sqrt{(-5)^2+4^2+(-7)^2} \\[6pt] &=& \sqrt{25+16+49} \\[6pt] &=& \sqrt{90}\\[12pt] \vert \overrightarrow{\textsf{AC}} \vert &=& \sqrt{4^2+(-7)^2+(-5)^2} \\[6pt] &=& \sqrt{16+49+25} \\[6pt] &=& \sqrt{90} \end{eqnarray} $$

Finally, we use the other definition of dot product to find angle BAC:

$$ \begin{gather} \overrightarrow{\textsf{AB}}\,.\,\overrightarrow{\textsf{AC}}=\vert \overrightarrow{\textsf{AB}} \vert\ \vert \overrightarrow{\textsf{AC}} \vert\ \text{cos}\,\textsf{BAC} \\[8pt] -13=\left(\sqrt{90}\,\right)\left(\sqrt{90}\,\right)\,\text{cos}\,\textsf{BAC} \\[8pt] -13=90\,\text{cos}\,\textsf{BAC} \\[8pt] \text{cos}\,\textsf{BAC}=-\frac{13}{90} \\[8pt] \end{gather} $$ $$ \begin{eqnarray} \angle\textsf{BAC} &=& \text{cos}^{-1}\ \left(-\frac{13}{90}\right) \\[8pt] &\approx& 98.3^\circ\:\small\textsf{or }\normalsize 1.72 \small\textsf{ radians} \\[8pt] \end{eqnarray} $$

There are three main ways to find a point that divides a line segment: 'stepping out', using basic vector algebra or using the 'sector formula'.

You may have been taught only one of these methods, or more than one. Use whichever you prefer.

However, bear in mind that this was only a 2-mark question, so whichever method you use, make sure you practise it enough, so that you can do it quickly and accurately, every time!

Method 1: Stepping out

\(x\!: -1-(-6)=5\small,\) which splits as \(2\!:\!3\small.\)
\(y\!: 11-1=10\small,\) which splits as \(4\!:\!6\small.\)
\(z\!: -8-2=-10\small,\) which splits as \(-4:-6\small.\)

Now we add each of these 'steps' (i.e. the first numbers in each ratio) onto the corresponding coordinates of P.

So R is \(\left(-6\!+\!2,\,1\!+\!4,\,2\!+(-\!4)\right)\) \(=(-4,\,5,\,-2).\)

Method 2: Basic vector algebra

$$ \begin{eqnarray} \overrightarrow{\textsf{PR}}:\overrightarrow{\textsf{RQ}} &=& 2:3 \\[6pt] 3\,\overrightarrow{\textsf{PR}} &=& 2\,\overrightarrow{\textsf{RQ}} \\[6pt] 3\,\left(\underline r - \underline p\right) &=& 2\,\left(\underline q - \underline r\right) \\[6pt] 3\underline r - 3\underline p &=& 2\underline q - 2\underline r \\[6pt] 3\underline r + 2\underline r &=& 2\underline q + 3\underline p \\[6pt] 5\underline r &=& 2\underline q + 3\underline p \\[6pt] &=& 2\left( \begin{matrix} -1\phantom{.} \\ \phantom{.}11\phantom{.} \\ -8\phantom{.} \end{matrix}\right) + 3\left( \begin{matrix} -6\phantom{.} \\ \phantom{-}1\phantom{.} \\ \phantom{-}2\phantom{.} \end{matrix}\right) \\[6pt] &=& \left( \begin{matrix} \phantom{.}-\!2\phantom{.} \\ \phantom{-}22\phantom{.} \\ -16\phantom{.} \end{matrix}\right) + \left( \begin{matrix} -18\phantom{.} \\ \phantom{-}3\phantom{.} \\ \phantom{-}6\phantom{.} \end{matrix}\right) \\[6pt] &=& \left( \begin{matrix} \,-20\phantom{.} \\ \,\phantom{-}25\phantom{.} \\ \,-10\phantom{.} \end{matrix}\right) \\[6pt] \underline r &=& \left( \begin{matrix} \,-4\, \\ \,\phantom{-}5\, \phantom{.} \\ -2\, \end{matrix}\right) \end{eqnarray} $$

So the coordinates of R are \((-4,\,5,\,-2).\)

Method 3: The section formula

We don't recommend this method – but if you like it, here it is!

Let \(m=2\) and \(n=3\small.\)

$$ \begin{eqnarray} \underline r &=& \frac{n\underline p + m\underline q}{n+m} \\[6pt] &=& \frac{3\underline p + 2\underline q}{2+3} \\[6pt] &=& \small\frac{1}{5}\normalsize\left(3\underline p + 2\underline q\right) \\[6pt] &=& \small\frac{1}{5}\left( \begin{matrix} 3(-6)+2(-1) \\ 3(1)+2(11) \\ 3(2)+2(-8) \end{matrix}\right) \\[6pt] &=& \small\frac{1}{5}\left( \begin{matrix} -18-2 \\ 3+22 \\ 6-16 \end{matrix}\right) \\[6pt] &=& \small\frac{1}{5}\left( \begin{matrix} \,-20\phantom{.} \\ \,\phantom{-}25\phantom{.} \\ \,-10\phantom{.} \end{matrix}\right) \\[6pt] &=& \left( \begin{matrix} \,-4\, \\ \,\phantom{-}5\, \\ \,-2\, \end{matrix}\right) \end{eqnarray} $$

So the coordinates of R are \((-4,\,5,\,-2)\small.\)

(a)   This 7-mark question broke down as 2 for (a), 1 for (b)(i) and 4 for (b)(ii).

$$ \begin{eqnarray} \overrightarrow{\textsf{ED}} &=& \overrightarrow{\textsf{OD}}-\overrightarrow{\textsf{OE}}\\[6pt] &=& \left( \begin{matrix} \phantom{-}2\phantom{.} \\ -3\phantom{.} \\ \phantom{-}4\phantom{.} \end{matrix}\right) - \left( \begin{matrix} \phantom{-}1\phantom{.} \\ \phantom{-}1\phantom{.} \\ -2\phantom{.} \end{matrix}\right)\\[6pt] &=& \left( \begin{matrix} \phantom{-}1\phantom{.} \\ -4\phantom{.} \\ \phantom{-}6\phantom{.} \end{matrix}\right)\\[15pt] \overrightarrow{\textsf{EF}} &=& \overrightarrow{\textsf{OF}}-\overrightarrow{\textsf{OE}}\\[6pt] &=& \left( \begin{matrix} \phantom{.}3\phantom{.} \\ \phantom{.}2\phantom{.} \\ \phantom{.}1\phantom{.} \end{matrix}\right) - \left( \begin{matrix} \phantom{-}1\phantom{.} \\ \phantom{-}1\phantom{.} \\ -2\phantom{.} \end{matrix}\right)\\[6pt] &=& \left( \begin{matrix} \phantom{.}2\phantom{.} \\ \phantom{.}1\phantom{.} \\ \phantom{.}3\phantom{.} \end{matrix}\right)\\[12pt] \end{eqnarray} $$

(b) (i)  We use the components to calculate the scalar product:

$$ \begin{eqnarray} \overrightarrow{\textsf{ED}}.\overrightarrow{\textsf{EF}} &=& -\!1(2)+(-4)(1)+(6)(3)\\[6pt] &=& -\!2-4+18\\[6pt] &=& \:16 \end{eqnarray} $$

(b) (ii)  To use the other definition of dot product, we need the magnitude of each vector:

$$ \begin{eqnarray} \vert\overrightarrow{\textsf{ED}}\vert &=& \sqrt{1^2+(-4)^2+6^2} \\[6pt] &=& \sqrt{1+16+36} \\[6pt] &=& \sqrt{53}\\[12pt] \vert\overrightarrow{\textsf{EF}}\vert &=& \sqrt{2^2+1^2+3^2} \\[6pt] &=& \sqrt{4+1+9} \\[6pt] &=& \sqrt{14} \end{eqnarray} $$

Now we are ready to find angle DEF:

$$ \begin{gather} \overrightarrow{\textsf{ED}}.\overrightarrow{\textsf{EF}}=\vert\overrightarrow{\textsf{ED}}\vert\,\vert\overrightarrow{\textsf{EF}}\vert\,\text{cos}\ \textsf{DEF}\normalsize \\[10pt] 16=\sqrt{53}\,\sqrt{14}\ \text{cos}\ \textsf{DEF}\normalsize \\[10pt] \text{cos}\ \textsf{DEF}=\frac{16}{\sqrt{53}\,\sqrt{14}} \\[10pt] \end{gather} $$ $$ \begin{eqnarray} \angle\textsf{DEF} &=& \text{cos}^{-1}\ \left(\frac{16}{\sqrt{53}\sqrt{14}}\right)\normalsize \\[10pt] &\approx& 54.03^\circ\:\small\textsf{or }\normalsize 0.943 \small\textsf{ radians} \\[10pt] \end{eqnarray} $$

This solution requires both definitions of scalar product, so first let's calculate it using the components.

$$ \begin{eqnarray} \underline{u}\,.\,\underline{v} &=& (1)(1)+(1)(3)+(0)(k)\\[6pt] &=& 1+3\\[6pt] &=& 4 \end{eqnarray} $$

We will also need the magnitude of each vector.

$$ \begin{eqnarray} \vert\underline{u}\vert &=& \sqrt{1^2+1+0^2} \\[6pt] &=& \sqrt{2}\\[15pt] \vert\underline{v}\vert &=& \sqrt{1^2+3^2+k^2} \\[6pt] &=& \sqrt{1+9+k^2} \\[6pt] &=& \sqrt{10+k^2} \end{eqnarray} $$

Finally, we use the other definition of scalar product and solve for \(k\small.\)

$$ \begin{gather} \underline{u}\,.\,\underline{v}=\vert\underline{u}\vert\ \vert\underline{v}\vert\ \text{cos}\,45^\circ \\[8pt] 4=\sqrt{2}\times\sqrt{10+k^2}\times\small\frac{1}{\sqrt{2}} \\[8pt] 4=\sqrt{10+k^2} \\[6pt] 16= 10+k^2 \\[6pt] k^2=6 \\[6pt] k=\sqrt{6} \\[6pt] \end{gather} $$

Note that we disregard the negative solution as the question tells us that \(k\gt 0\small.\)

(a)  Because B divides AC, it makes sense to choose two vectors that both involve B.

$$ \begin{eqnarray} \overrightarrow{\textsf{AB}} &=& \overrightarrow{\textsf{OB}}-\overrightarrow{\textsf{OA}}\\[6pt] &=& \left( \begin{matrix} \phantom{-}6\, \\ -1\, \\ \phantom{-}5\, \end{matrix}\right) - \left( \begin{matrix} -3\, \\ \phantom{-}2\, \\ -9\, \end{matrix}\right)\\[6pt] &=& \left( \begin{matrix} \phantom{-}9\, \\ -3\, \\ \phantom{-}6\, \end{matrix}\right)\\[15pt] \overrightarrow{\textsf{BC}} &=& \overrightarrow{\textsf{OC}} - \overrightarrow{\textsf{OB}}\\[6pt] &=& \left( \begin{matrix} 12\, \\ -3\, \\ \phantom{-}9\, \end{matrix}\right) - \left( \begin{matrix} \phantom{-}6\, \\ -1\, \\ \phantom{-}5\, \end{matrix}\right)\\[6pt] &=& \left( \begin{matrix} \phantom{-}6\, \\ -2\, \\ \phantom{-}4\, \end{matrix}\right) \\[12pt] \end{eqnarray} $$

From the above, \(\overrightarrow{\textsf{AB}}=\large\frac{3}{2}\normalsize\,\overrightarrow{\textsf{BC}}\) so \(\overrightarrow{\textsf{AB}}\) and \(\overrightarrow{\textsf{BC}}\) are parallel. B is a common point, so A, B and C are collinear.

(b)  With experience, you can usually write down these ratios without any further working, but a more formal method would be to start with the connection between \(\overrightarrow{\textsf{AB}}\) and \(\overrightarrow{\textsf{BC}}\) from part (a) above:

$$ \begin{gather} \overrightarrow{\textsf{AB}}=\frac{3}{2}\,\overrightarrow{\textsf{BC}} \\[8pt] \frac{\overrightarrow{\textsf{AB}}}{\overrightarrow{\textsf{BC}}}=\frac{3}{2} \\[8pt] \overrightarrow{\textsf{AB}}:\overrightarrow{\textsf{BC}}=3:2 \end{gather} $$

First, we need to decide a route from B to E, along edges whose vectors we know.

\(\overrightarrow{\textsf{DE}}\) is the only given vector that can take us to E, so we will need to get from B to D first.

One suitable route is therefore: B to A, then to D, and finally to E. So:

$$ \begin{eqnarray} \overrightarrow{\textsf{BE}} &=& \overrightarrow{\textsf{BA}}+\overrightarrow{\textsf{AD}}+\overrightarrow{\textsf{DE}} \\[6pt] &=& -\!\overrightarrow{\textsf{DC}}+\overrightarrow{\textsf{AD}}+\overrightarrow{\textsf{DE}} \\[6pt] &=& -2\underline{i}+4\underline{j}-2\underline{k} \\[6pt] && +6\underline{i}+4\underline{j}+2\underline{k}\\[6pt] && -4\underline{i}-3\underline{j}+4\underline{k}\\[6pt] &=& \:\:5\underline{j}+4\underline{k} \end{eqnarray} $$