2020 H2 Maths Paper 1 Q5 — Worked Solution

Part (a)

2 marks

Since the cross product is anti-commutative, $\mathbf{b} \times \mathbf{a} = - \left( \mathbf{a} \times \mathbf{b} \right)$ for any vectors $\mathbf{a}$ and $\mathbf{b}$ . Substituting the given identity $\mathbf{a} \times \mathbf{b} = \mathbf{b} \times \mathbf{a}$ :

\mathbf{a} \times \mathbf{b} = - \left( \mathbf{a} \times \mathbf{b} \right)

\Rightarrow 2\left( \mathbf{a} \times \mathbf{b} \right) = \mathbf{0}

\Rightarrow \mathbf{a} \times \mathbf{b} = \mathbf{0}

Since $\mathbf{a}$ and $\mathbf{b}$ are both non-zero, $\mathbf{a} \times \mathbf{b} = \mathbf{0}$ means $\mathbf{a}$ and $\mathbf{b}$ must be parallel.

To check your answer: if

\mathbf{a}

and

\mathbf{b}

are parallel, say

\mathbf{b} = k\mathbf{a}

for some scalar

k

, then

\mathbf{a} \times \mathbf{b} = \mathbf{a} \times \left( k\mathbf{a} \right) = k\left( \mathbf{a} \times \mathbf{a} \right) = \mathbf{0}

, and likewise

\mathbf{b} \times \mathbf{a} = \mathbf{0}

, so indeed

\mathbf{a} \times \mathbf{b} = \mathbf{b} \times \mathbf{a}

. ✓

$\therefore$ $\mathbf{a}$ and $\mathbf{b}$ are parallel.

Part (b)(i)

3 marks

Since $\mathbf{q}$ is non-zero, the condition $\left( \mathbf{r} - \mathbf{p} \right) \times \mathbf{q} = \mathbf{0}$ means, by the same reasoning as part (a), that $\overrightarrow{PR} = \mathbf{r} - \mathbf{p}$ is parallel to $\mathbf{q}$ .

$\therefore$ $R$ lies on the straight line through $P$ with direction vector $\mathbf{q}$ .

Part (b)(ii)

4 marks

Using the given position vectors,

\mathbf{r} - \mathbf{p} = \begin{pmatrix} x - ( - 1) \\ y - 2 \\ z - 4 \end{pmatrix} = \begin{pmatrix} x + 1 \\ y - 2 \\ z - 4 \end{pmatrix}

Forming the dot product with $\mathbf{q}$ ,

\left( \mathbf{r} - \mathbf{p} \right) \cdot \mathbf{q}

= 3(x + 1) - 5(y - 2) + 2(z - 4)

= 3x + 3 - 5y + 10 + 2z - 8

= 3x - 5y + 2z + 5

Setting this equal to $0$ ,

3x - 5y + 2z + 5 = 0

\Rightarrow 3x - 5y + 2z = - 5

To check your answer: substituting

P = ( - 1,2,4)

into the equation,

3( - 1) - 5(2) + 2(4) = - 3 - 10 + 8 = - 5

, matching the right side — as expected, since a plane through

P

must contain

P

. ✓

Since $\left( \mathbf{r} - \mathbf{p} \right) \cdot \mathbf{q} = 0$ means $\overrightarrow{PR}$ is perpendicular to $\mathbf{q}$ , the point $R$ lies on the plane through $P$ with normal vector $\mathbf{q}$ .

$\therefore$ $3x - 5y + 2z = - 5$ ; $R$ lies on the plane through $P$ with normal vector $\mathbf{q}$ .

2020 H2 Maths Paper 1, Question 5

Part (a)

Part (b)(i)

Part (b)(ii)

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