Let a=i+2 j+3 k and b=i+j-k. If c is a vector such that a c=11, b… - Vidyadip

MCQ

Let $\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}$ and $\vec{b}=\hat{i}+\hat{j}-\hat{k}$. If $\vec{c}$ is a vector such that $\vec{a} \cdot \vec{c}=11, \vec{b} \cdot(\vec{a} \times \vec{c})=27$ and $\vec{b} \cdot \vec{c}=-\sqrt{3}|\vec{b}|$, then $|\vec{a} \times \vec{c}|^2$ is equal to

✓
$285$
B
$284$
C
$283$
D
$282$

✓

Answer

Correct option: A.

$285$

a
Sol. $\overrightarrow{ a }=\hat{ i }+2 \hat{ j }+3 \hat{ k }, \overrightarrow{ b }=\hat{ i }+\hat{ j }-\hat{ k }$

$\vec{b} \cdot(\vec{a} \times \vec{c})=27, \vec{a} \cdot \vec{b}=0$

$\vec{b} \times(\vec{a} \times \vec{c})=-3 \vec{a}$

Let $\theta$ be angle between $\vec{b}, \vec{a} \times \vec{c}$

Then $|\vec{b}| \cdot|\vec{a} \times \vec{c}| \sin \theta=3 \sqrt{14}$

$|\vec{b}| \cdot|\vec{a} \times \vec{c}| \cos \theta=27$

$\Rightarrow \sin \theta=\frac{\sqrt{14}}{\sqrt{95}}$

$\therefore|\overrightarrow{ b }| \times|\overrightarrow{ a } \times \overrightarrow{ c }|=3 \sqrt{95}$

$\Rightarrow|\overrightarrow{ a } \times \overrightarrow{ c }|=\sqrt{3} \times \sqrt{95}$

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