2016 H2 Maths Paper 1 Q5 — Worked Solution

Part (i)

6 marks

Approach

Expanding $\left( \mathbf{u} + \mathbf{v} \right) \times \left( \mathbf{u} - \mathbf{v} \right)$ directly by the distributive law is quicker than computing $\mathbf{u} \times \mathbf{v}$ first and doubling it, since the cross product is not commutative and the signs are easy to drop; we do it directly with components here.

Since $\mathbf{u} = 2\mathbf{i} - \mathbf{j} + 2\mathbf{k}$ and $\mathbf{v} = a\mathbf{i} + b\mathbf{k}$ ,

\mathbf{u} + \mathbf{v} = (2 + a)\mathbf{i} - \mathbf{j} + (2 + b)\mathbf{k},\quad\quad\mathbf{u} - \mathbf{v} = (2 - a)\mathbf{i} - \mathbf{j} + (2 - b)\mathbf{k}.

Taking the cross product, with $\mathbf{u} + \mathbf{v} = (2 + a)\mathbf{i} - \mathbf{j} + (2 + b)\mathbf{k}$ and $\mathbf{u} - \mathbf{v} = (2 - a)\mathbf{i} - \mathbf{j} + (2 - b)\mathbf{k}$ , each component of the cross product is found from the other two components of the two vectors:

\mathbf{i}\text{-component}:( - 1)(2 - b) - (2 + b)( - 1) = - 2 + b + 2 + b = 2b

\mathbf{j}\text{-component}: - \left\lbrack (2 + a)(2 - b) - (2 + b)(2 - a) \right\rbrack = - ( - 4a + 4b) = 4a - 4b

\mathbf{k}\text{-component}:(2 + a)( - 1) - ( - 1)(2 - a) = - 2 - a + 2 - a = - 2a

\therefore\ \left( \mathbf{u} + \mathbf{v} \right) \times \left( \mathbf{u} - \mathbf{v} \right) = 2b\mathbf{i} + (4a - 4b)\mathbf{j} - 2a\mathbf{k}.

Part (ii)

Using the given probability, the $\mathbf{i}$ - and $\mathbf{k}$ -components of the part (i) answer are equal:

2b = - 2a

b = - a

Substituting $b = - a$ into the part (i) answer,

\left( \mathbf{u} + \mathbf{v} \right) \times \left( \mathbf{u} - \mathbf{v} \right) = 2( - a)\mathbf{i} + \left( 4a - 4( - a) \right)\mathbf{j} - 2a\mathbf{k}

\therefore\ \left( \mathbf{u} + \mathbf{v} \right) \times \left( \mathbf{u} - \mathbf{v} \right) = - 2a\mathbf{i} + 8a\mathbf{j} - 2a\mathbf{k}.

Approach

For this vector to be a unit vector its magnitude must equal $1$ . Since every component is a multiple of $a$ , factor $a$ out of the magnitude first.

\left| \left( \mathbf{u} + \mathbf{v} \right) \times \left( \mathbf{u} - \mathbf{v} \right) \right| = \sqrt{( - 2a)^{2} + (8a)^{2} + ( - 2a)^{2}} = \sqrt{72a^{2}} = 6\sqrt{2}\,|a|

Setting this equal to $1$ ,

6\sqrt{2}\,|a| = 1

|a| = \frac{1}{6\sqrt{2}} = \frac{\sqrt{2}}{12}

To check your answer: with

a = \frac{\sqrt{2}}{12}

,

6\sqrt{2} \cdot \frac{\sqrt{2}}{12} = \frac{12}{12} = 1

. ✓

$\therefore$ $a = \frac{\sqrt{2}}{12}$ or $a = - \frac{\sqrt{2}}{12}$ .

Part (iii)

2 marks

Approach

Expanding $\left( \mathbf{u} + \mathbf{v} \right).\left( \mathbf{u} - \mathbf{v} \right)$ by the distributive law of the dot product, the cross terms $- \mathbf{u}.\mathbf{v}$ and $\text{+}\mathbf{v}.\mathbf{u}$ cancel since the dot product is commutative, leaving a relation between $\left| \mathbf{u} \right|^{2}$ and $\left| \mathbf{v} \right|^{2}$ only — so the individual values of $a$ and $b$ are never needed.

\left( \mathbf{u} + \mathbf{v} \right).\left( \mathbf{u} - \mathbf{v} \right)

= \mathbf{u}.\mathbf{u} - \mathbf{u}.\mathbf{v} + \mathbf{v}.\mathbf{u} - \mathbf{v}.\mathbf{v}

= \left| \mathbf{u} \right|^{2} - \left| \mathbf{v} \right|^{2}

Using the given condition $\left( \mathbf{u} + \mathbf{v} \right).\left( \mathbf{u} - \mathbf{v} \right) = 0$ ,

\left| \mathbf{u} \right|^{2} - \left| \mathbf{v} \right|^{2} = 0

\left| \mathbf{v} \right|^{2} = \left| \mathbf{u} \right|^{2}

Since $\left| \mathbf{u} \right|^{2} = 2^{2} + ( - 1)^{2} + 2^{2} = 9$ ,

\left| \mathbf{v} \right|^{2} = 9

As $\left| \mathbf{v} \right|$ is a length, $\left| \mathbf{v} \right| > 0$ , so we reject the negative root.

$\therefore$ $\left| \mathbf{v} \right| = 3$ .

2016 H2 Maths Paper 1, Question 5

Part (i)

Part (ii)

Part (iii)

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