Vectors 2 Question 11
11. Let $\triangle P Q R$ be a triangle. Let $\mathbf{a}=\mathbf{Q R}, \mathbf{b}=\mathbf{R P}$ and $\mathbf{c}=\mathbf{P Q}$. If $|\mathbf{a}|=12,|\mathbf{b}|=4 \sqrt{3}$ and $\mathbf{b} \cdot \mathbf{c}=24$, then which of the following is/are true?
(2015 Adv.)
(a) $\frac{|\mathbf{c}|^{2}}{2}-|\mathbf{a}|=12$
(b) $\frac{|\mathbf{c}|^{2}}{2}+|\mathbf{a}|=30$
(c) $|\mathbf{a} \times \mathbf{b}+\mathbf{c} \times \mathbf{a}|=48 \sqrt{3}$
(d) $\mathbf{a} \cdot \mathbf{b}=-72$
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Answer:
Correct Answer: 11. (a, c, d)
Solution:
- Given, $|\mathbf{a}|=12,|\mathbf{b}|=4 \sqrt{3}$
$ \begin{aligned} & \mathbf{a}+\mathbf{b}+\mathbf{c}=\mathbf{0} \\ & \Rightarrow \quad \mathbf{a}=-(\mathbf{b}+\mathbf{c}) \\ & \text { We have, } \quad|\mathbf{a}|^{2}=|\mathbf{b}+\mathbf{c}|^{2} \\ & \Rightarrow \quad|\mathbf{a}|^{2}=|\mathbf{b}|^{2}+|\mathbf{c}|^{2}+2 \mathbf{b} \cdot \mathbf{c} \\ & \Rightarrow \quad 144=48+|\mathbf{c}|^{2}+48 \\ & \Rightarrow \quad|\mathbf{c}|^{2}=48 \\ & \Rightarrow \quad|\mathbf{c}|=4 \sqrt{3} \\ & \text { Also, } \quad|\mathbf{c}|^{2}=|\mathbf{a}|^{2}+|\mathbf{b}|^{2}+2 \mathbf{a} \cdot \mathbf{b} \\ & \Rightarrow \quad 48=144+48+2 \mathbf{a} \cdot \mathbf{b} \\ & \Rightarrow \quad \mathbf{a} \cdot \mathbf{b}=-72 \end{aligned} $
$\therefore$ Option (d) is correct.
Also,
$ \mathbf{a} \times \mathbf{b}=\mathbf{c} \times \mathbf{a} $
$\Rightarrow \mathbf{a} \times \mathbf{b}+\mathbf{c} \times \mathbf{a}=2 \mathbf{a} \times \mathbf{b}$
$\Rightarrow|\mathbf{a} \times \mathbf{b}+\mathbf{c} \times \mathbf{a}|=2|\mathbf{a} \times \mathbf{b}|=2 \sqrt{|\mathbf{a}|^{2}|\mathbf{b}|^{2}-(\mathbf{a} \cdot \mathbf{b})^{2}}$
$=2 \sqrt{(144)(48)-(-72)^{2}}$
$=2(12) \sqrt{48-36}=48 \sqrt{3}$
$\therefore$ Option (c) is correct.
Also, $\quad \frac{|\mathbf{c}|^{2}}{2}-|\mathbf{a}|=24-12=12$
$\therefore$ Option (a) is correct.
and
$ \frac{|\mathbf{c}|^{2}}{2}+|\mathbf{a}|=24+12=36 $
$\therefore$ Option (b) is not correct.
