Q74 / 160 · TS EAMCET 2016 · Mathematics
$\mathbf{a}$ is perpendicular to both $\mathbf{b}$ and $\mathbf{c}$. The angle between $\mathbf{b}$ and $\mathbf{c}$ is $\frac{2 \pi}{3}$. If $|\mathbf{a}|=2$, $|\mathbf{b}|=3,|\mathbf{c}|=4$, then $\mathbf{c} \cdot(\mathbf{a} \times \mathbf{b})$ is equal to
- A$18 \sqrt{3}$
- Correct answerB$12 \sqrt{3}$
- C$8 \sqrt{3}$
- D$6 \sqrt{3}$
Solution
Given,
$\begin{array}{rlrl} <br />& \mathbf{a} \cdot \mathbf{b}=0 \\<br />& \mathbf{a} \cdot \mathbf{c}=0 \\<br />& \mathbf{a}|| \mathbf{b} \times \mathbf{c} \\<br />\text { and } & |\mathbf{b} \times \mathbf{c}|=|\mathbf{b}| \cdot|\mathbf{c}| \sin \left(\frac{2 \pi}{3}\right) \\<br />& & {[\because|\mathbf{a}|=2,|\mathbf{b}|=3 \text { and }|\mathbf{c}|=4]} \\<br />& =3 \times 4 \times \frac{\sqrt{3}}{2}=6 \sqrt{3}<br />\end{array}$
and
$\begin{aligned}<br />& |\mathbf{b} \times \mathbf{c}|=|\mathbf{b}| \cdot|\mathbf{c}| \sin \left(\frac{2 \pi}{3}\right) \\<br />& {[\because|\mathbf{a}|=2,|\mathbf{b}|=3 \text { and }|\mathbf{c}|=4]} \\<br />& =3 \times 4 \times \frac{\sqrt{3}}{2}=6 \sqrt{3}<br />\end{aligned}$
Hence,
$\begin{aligned}<br />\mathbf{c} & \cdot(\mathbf{a} \times \mathbf{b})=\mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})[\because a \cdot b=|a||b| \cos \theta] \\<br />& =|\mathbf{a}||\mathbf{b} \times \mathbf{c}|=2 \times 6 \sqrt{3} \\<br />& =12 \sqrt{3}<br />\end{aligned}$
$\begin{array}{rlrl} <br />& \mathbf{a} \cdot \mathbf{b}=0 \\<br />& \mathbf{a} \cdot \mathbf{c}=0 \\<br />& \mathbf{a}|| \mathbf{b} \times \mathbf{c} \\<br />\text { and } & |\mathbf{b} \times \mathbf{c}|=|\mathbf{b}| \cdot|\mathbf{c}| \sin \left(\frac{2 \pi}{3}\right) \\<br />& & {[\because|\mathbf{a}|=2,|\mathbf{b}|=3 \text { and }|\mathbf{c}|=4]} \\<br />& =3 \times 4 \times \frac{\sqrt{3}}{2}=6 \sqrt{3}<br />\end{array}$
and
$\begin{aligned}<br />& |\mathbf{b} \times \mathbf{c}|=|\mathbf{b}| \cdot|\mathbf{c}| \sin \left(\frac{2 \pi}{3}\right) \\<br />& {[\because|\mathbf{a}|=2,|\mathbf{b}|=3 \text { and }|\mathbf{c}|=4]} \\<br />& =3 \times 4 \times \frac{\sqrt{3}}{2}=6 \sqrt{3}<br />\end{aligned}$
Hence,
$\begin{aligned}<br />\mathbf{c} & \cdot(\mathbf{a} \times \mathbf{b})=\mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})[\because a \cdot b=|a||b| \cos \theta] \\<br />& =|\mathbf{a}||\mathbf{b} \times \mathbf{c}|=2 \times 6 \sqrt{3} \\<br />& =12 \sqrt{3}<br />\end{aligned}$
Hence, the correct answer is option: 'B'.