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

Detailed breakdown by ChatGPT 

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:

Detailed breakdown by ChatGPT 

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.

Detailed breakdown by ChatGPT 

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.

Detailed breakdown by ChatGPT 

(a)  As part (b) tells us that 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:

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

Detailed breakdown by ChatGPT 

(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:

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

Detailed breakdown by ChatGPT 

There are three main methods to divide a line segment in a given ratio.

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 and rearranging to find the position vector of S.

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

Method 3: The section formula

$$ \underline s=\frac{n\underline r + m\underline t}{n+m} $$

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 \((3,\,1,\,2).\)

Detailed breakdown by ChatGPT 

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 \(k=\large\frac{1}{9}\small.\)

Detailed breakdown by ChatGPT 

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 \(\text{cos}\,90^\circ=0\small.\)

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

Detailed breakdown by ChatGPT 

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:

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

Detailed breakdown by ChatGPT 

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

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

Detailed breakdown by ChatGPT 

(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 \(\theta\small.\)

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

Detailed breakdown by ChatGPT 

(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:

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

Detailed breakdown by ChatGPT 

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.

Detailed breakdown by ChatGPT 

(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 \((9,\,-8,\,5)\small.\)

Detailed breakdown by ChatGPT 

(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:

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

Detailed breakdown by ChatGPT 

There are three main ways to find a point that divides a line segment.

This was only a 2-mark question, so whichever method you use, make sure you can do it quickly and accurately!

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

Let \(m=2\) and \(n=3\small.\) The section formula is:

$$ \underline r=\frac{n\underline p + m\underline q}{n+m} $$

We don't recommend this method, but it is accepted in exams, so here is the working:

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

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

Detailed breakdown by ChatGPT 

(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)  First 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 can find angle DEF:

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

Detailed breakdown by ChatGPT 

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 \(k\gt 0\small.\)

Detailed breakdown by ChatGPT 

(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:

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

Detailed breakdown by ChatGPT 

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:

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

Detailed breakdown by ChatGPT 

(a)  The method here is to show that two vectors with a common point are parallel. For example:

$$ \begin{eqnarray} \overrightarrow{\textsf{DE}} &=& \overrightarrow{\textsf{OE}}-\overrightarrow{\textsf{OD}}\\[6pt] &=& \left( \begin{matrix} \!\phantom{-}1\, \\ \!\phantom{-}2\, \\ \!\phantom{-}7\, \end{matrix}\right) - \left( \begin{matrix} -1\, \\ \phantom{-}6\, \\ \phantom{-}1\, \end{matrix}\right)\\[6pt] &=& \left( \begin{matrix} \phantom{-}2\, \\ -4\, \\ \phantom{-}6\, \end{matrix}\right)\\[12pt] \overrightarrow{\textsf{EF}} &=& \overrightarrow{\textsf{OF}}-\overrightarrow{\textsf{OE}}\\[6pt] &=& \left( \begin{matrix} \phantom{-}2\, \\ \phantom{-}0\, \\ \phantom{.}10\, \end{matrix}\right) - \left( \begin{matrix} \!\phantom{-}1\, \\ \!\phantom{-}2\, \\ \!\phantom{-}7\, \end{matrix}\right)\\[6pt] &=& \left( \begin{matrix} \phantom{-}1\, \\ -2\, \\ \phantom{-}3\, \end{matrix}\right) \\[12pt] \end{eqnarray} $$

From the above, \(\overrightarrow{\textsf{DE}}=2\,\overrightarrow{\textsf{EF}}\small,\) so \(\overrightarrow{\textsf{DE}}\) and \(\overrightarrow{\textsf{EF}}\) are parallel.

E is a common point, so D, E and F are collinear.

(b)  This question includes a little twist, because G is one of the endpoints, not the internal division point.

The easiest method is probably to find \(\overrightarrow{\textsf{DF}}\small,\) use the given ratio to find \(\overrightarrow{\textsf{FG}}\) and then use that to find G.

$$ \begin{eqnarray} \overrightarrow{\textsf{DF}} &=& \overrightarrow{\textsf{OF}}-\overrightarrow{\textsf{OD}}\\[6pt] &=& \left( \begin{matrix} \!\phantom{-}2\, \\ \!\phantom{-}0\, \\ \!\phantom{.}10\, \end{matrix}\right) - \left( \begin{matrix} -1\, \\ \phantom{-}6\, \\ \phantom{-}1\, \end{matrix}\right)\\[6pt] &=& \left( \begin{matrix} \phantom{-}3\, \\ -6\, \\ \phantom{-}9\, \end{matrix}\right)\\[12pt] \end{eqnarray} $$

Now, because DF:FG = \(3\!:\!4\small,\) we can just divide each component of \(\overrightarrow{\textsf{DF}}\) by 3 and multiply by 4. So:

$$ \overrightarrow{\textsf{FG}} = \left( \begin{matrix} \phantom{-}4\, \\ -8\, \\ \phantom{.}12\, \end{matrix}\right) $$

All that remains is to add the components of \(\overrightarrow{\textsf{FG}}\) to the coordinates of F.

So G is \(\,(2+4,\,0-8,\,10+12)\) which is \((6,\,-8,\,22)\small.\)

Video solution by Clelland Maths 
Detailed breakdown by ChatGPT 

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

$$ \begin{eqnarray} \underline{u}\,.\,\underline{v} &=& 3\!\times\!(-5) + 2\!\times\!(-2) + 6\!\times\!4\\[6pt] &=& -\!15-4+24\\[6pt] &=& 5 \end{eqnarray} $$

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

$$ \begin{eqnarray} \vert\underline{u}\vert &=& \sqrt{3^2+2^2+6^2} \\[6pt] &=& \sqrt{9+4+36} \\[6pt] &=& \sqrt{49} \\[6pt] &=& 7\\[15pt] \vert\underline{v}\vert &=& \sqrt{(-5)^2+(-2)^2+4^2} \\[6pt] &=& \sqrt{25+4+16} \\[6pt] &=& \sqrt{45} \end{eqnarray} $$

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

$$ \begin{gather} \underline{u}\,.\,\underline{v}=\vert\underline{u}\vert\ \vert\underline{v}\vert\ \text{cos}\,\theta \\[8pt] 5=7\,\sqrt{45}\,\text{cos}\,\theta \\[8pt] \text{cos}\,\theta=\frac{5}{7\sqrt{45}} \\[8pt] \end{gather} $$ $$ \begin{eqnarray} \theta &=& \text{cos}^{-1}\ \frac{5}{7\sqrt{45}} \\[8pt] &=& 83.8875... \\[8pt] &\approx& 83.9^\circ\:\small\textsf{(to 1 d.p.)} \\[8pt] \end{eqnarray} $$

Detailed breakdown by ChatGPT