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Vectors: EJC 2024 Promo Q5

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Eunoia Junior College2024 PromoPromo Q5●●○ Standard6 marks

What this question tests

Vectors (Products).

Question

$\mathbf{a}$ and $\mathbf{b}$ are vectors such that $|\mathbf{a}| = \sqrt{2}$ and $|\mathbf{b}| = \sqrt{6}$ . The angle between $\mathbf{a}$ and $\mathbf{b}$ is $\dfrac{5\pi}{6}$ .

Find the exact length of projection of $\mathbf{a}$ onto $\mathbf{b}$ . [2]
By considering $(3\mathbf{a} + 2\mathbf{b})\cdot(3\mathbf{a} + 2\mathbf{b})$ , find the exact value of $|3\mathbf{a} + 2\mathbf{b}|$ . [4]

Show full worked solution▾

(a)

Required length of projection is $\begin{aligned} \left|\,|\mathbf{a}|\cos\frac{5\pi}{6}\,\right| &= \left|\sqrt{2}\times\left(-\frac{\sqrt{3}}{2}\right)\right| = \frac{\sqrt{6}}{2}. \end{aligned}$

(b)

$\begin{aligned} |3\mathbf{a} + 2\mathbf{b}|^2 &= (3\mathbf{a} + 2\mathbf{b})\cdot(3\mathbf{a} + 2\mathbf{b})\\ &= (3\mathbf{a})\cdot(3\mathbf{a})+(2\mathbf{b})\cdot(3\mathbf{a}) +(3\mathbf{a})\cdot(2\mathbf{b})+(2\mathbf{b})\cdot(2\mathbf{b})\\ &= 9|\mathbf{a}|^2 + 12\mathbf{a}\cdot\mathbf{b} + 4|\mathbf{b}|^2\\ &= 9(2) + 12(\sqrt{2})(\sqrt{6})\left(-\frac{\sqrt{3}}{2}\right) + 4(6)\\ &= 6. \end{aligned}$ So $|3\mathbf{a} + 2\mathbf{b}| = \sqrt{6}$ .

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