Higher Maths
Vectors

(a)  For \(\small\overrightarrow{\textsf{RT}}\normalsize,\) we imagine starting at R and going via S to T.

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

(b)  For \(\small\overrightarrow{\textsf{RP}}\normalsize,\) we imagine starting at R and going via T to P.

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

(a)  Imagine travelling from R to P via Q:

(b)  Let's go from R to S via P:

$$ \begin{eqnarray} \small \overrightarrow{\textsf{RS}} \normalsize &=& \small\overrightarrow{\textsf{RP}} \normalsize + \small \overrightarrow{\textsf{PS}} \normalsize \\[6pt] &=& \small\overrightarrow{\textsf{RP}} \normalsize - \small\frac{1}{2}\small \overrightarrow{\textsf{RQ}} \:\:\:\:\small\textsf{(because } \small \overrightarrow{\textsf{RQ}} \normalsize = 2\tiny\ \small \overrightarrow{\textsf{SP}} \normalsize \textsf{)} \\[6pt] &=& (\underline v - \underline u) - \small\frac{1}{2} \normalsize \underline v \:\:\:\:\small\textsf{(using part (a))}\\[6pt] &=& \small\frac{1}{2}\normalsize\underline v - \underline u \end{eqnarray} $$

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

\(\small\overrightarrow{\textsf{AC}}\normalsize=-2\small\overrightarrow{\textsf{AB}}\normalsize\) so \(\small\overrightarrow{\textsf{AC}}\normalsize\) and \(\small\overrightarrow{\textsf{AB}}\normalsize\) 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.

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

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

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

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

\(\small\overrightarrow{\textsf{FG}}\normalsize=\frac{3}{2}\small\overrightarrow{\textsf{GH}}\normalsize\) so \(\small\overrightarrow{\textsf{FG}}\normalsize\) and \(\small\overrightarrow{\textsf{GH}}\normalsize\) 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 \(\small\overrightarrow{\textsf{FG}}\normalsize\) and \(\small\overrightarrow{\textsf{GH}}\normalsize\) from part (a) above:

(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:

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

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

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

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

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

Note: We don't like the section formula and won't use it in this example. It might be a bit quicker, but it's ugly, it's easy to forget and it discourages logical process. We much prefer this method:

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

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

Ugh! No! The section formula is horrible! Well, OK, if we must...

First a reminder of the question:

Question: R and T are the points \((7,-1,8)\) and \(\small\textsf{C}\normalsize(-3,4,-7)\) respectively. Point S divides RT internally in the ratio \(2:3.\) Determine the coordinates of point S.

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

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

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

Note: We absolutely hate the section formula – but whether you agree or disagree, the good news is that the SQA won't tell you what method to use for questions like this. Use what suits you best.

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

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

Now, as \(\vert \small\overrightarrow{\textsf{AB}}\normalsize \vert =9\) and \(\vert k\,\small\overrightarrow{\textsf{AB}}\normalsize \vert =1,\) we can conclude that \(k=\frac{1}{9}.\)

Questions to do with the angle between vectors always use the scalar product, as defined at the top of this page. The two definitions are included on the formulae list in your exams.

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

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

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

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

Now we need the scalar product:

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

Next, we need the magnitude of each vector:

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

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

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

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

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

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

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

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

$$ \begin{eqnarray} \vert\underline{u}\vert &=& \sqrt{(-1)^2+4^2+(-3)^2} \\[6pt] &=& \sqrt{1+16+9} \\[6pt] &=& \sqrt{26}\\[12pt] \end{eqnarray} $$

$$ \begin{eqnarray} \vert\underline{v}\vert &=& \sqrt{(-7)^2+8^2+5^2} \\[6pt] &=& \sqrt{49+64+25} \\[6pt] &=& \sqrt{138} \end{eqnarray} $$

Finally, we calculate the angle. Let's call it \(\theta\small.\)

$$ \begin{eqnarray} \underline{u}.\underline{v} &=& \vert\underline{u}\vert\ \vert\underline{v}\vert\ \small\textsf{cos}\normalsize\,\theta \\[6pt] 24 &=& \sqrt{26}\sqrt{138}\,\small\textsf{cos}\normalsize\,\theta \\[6pt] \small\textsf{cos}\normalsize\,\theta &=& \small\frac{24}{\sqrt{26}\sqrt{138}}\normalsize \\[6pt] \theta &=& \small\textsf{cos}^{-1}\ \frac{24}{\sqrt{26}\sqrt{138}}\normalsize \\[6pt] &\approx& 66.4^\circ\:\small\textsf{(to 1 d.p.)}\normalsize \\[6pt] \end{eqnarray} $$

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

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

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

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

(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 \(-3=\frac32\!\times\!\!-\!2\) and \(6=\frac32\!\times\!4\small.\)

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

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

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

We start by expanding the bracket and using the fact that the angle between \(\underline{u}\) and itself is \(0^\circ.\)

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

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

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

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

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

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 \(\small\overrightarrow{\textsf{AC}}\normalsize=\frac{5}{3}\small\,\overrightarrow{\textsf{AB}}.\)

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

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

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

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

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

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

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

Next, we find the magnitude of each vector:

$$ \begin{eqnarray} \vert \small\overrightarrow{\textsf{AB}}\normalsize \vert &=& \sqrt{(-5)^2+4^2+(-7)^2} \\[6pt] &=& \sqrt{25+16+49} \\[6pt] &=& \sqrt{90}\\[9pt] \vert \small\overrightarrow{\textsf{AC}}\normalsize \vert &=& \sqrt{4^2+(-7)^2+(-5)^2} \\[6pt] &=& \sqrt{16+49+25} \\[6pt] &=& \sqrt{90} \end{eqnarray} $$

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

$$ \begin{eqnarray} \small\overrightarrow{\textsf{AB}}\normalsize . \small\overrightarrow{\textsf{AC}}\normalsize &=& \vert \small\overrightarrow{\textsf{AB}}\normalsize \vert\ \vert \small\overrightarrow{\textsf{AC}}\normalsize \vert\ \small\textsf{cos BAC}\normalsize \\[6pt] -13 &=& (\sqrt{90})(\sqrt{90})\,\small\textsf{cos BAC}\normalsize \\[6pt] -13 &=& 90\,\small\textsf{cos BAC}\normalsize \\[6pt] \small\textsf{cos BAC}\normalsize &=& -\small\frac{13}{90}\normalsize \\[6pt] \small\textsf{BAC}\normalsize &=& \small\textsf{cos}^{-1}\ \left(-\frac{13}{90}\right)\normalsize \\[6pt] &\approx& 98.3^\circ\:\small\textsf{or }\normalsize 1.72 \small\textsf{ radians} \\[6pt] \end{eqnarray} $$