Vector Algebra Question 19
Question 19 - 31 January - Shift 2
Let $\vec{a}, \vec{b}, \vec{c}$ be three vectors such that $|\overrightarrow{{}a}|=\sqrt{31}, 4|\overrightarrow{{}b}|=|\overrightarrow{{}c}|=2$ and $2(\overrightarrow{{}a} \times \overrightarrow{{}b})=3(\overrightarrow{{}c} \times \overrightarrow{{}a})$. If the angle between $\vec{b}$ and $\vec{c}$ is $\frac{2 \pi}{3}$, then $(\frac{\vec{a} \times \vec{c}}{\vec{a} \cdot \vec{b}})^{2}$ is equal to
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Answer: 3
Solution:
Formula: Vector product of two vectors when two vectors are parallel, Vector products of two vectors: Lagrange’s identity
b $3 \overrightarrow{{}c} \times \overrightarrow{{}a}$
$ \begin{aligned} & \vec{a} \times(2 \vec{b}+3 \vec{c})=0 \\ & \vec{a}=\lambda(2 \vec{b}+3 \vec{c}) \end{aligned} $
$|\overrightarrow{{}a}|^{2}=\lambda^{2}|2 \overrightarrow{{}b}+3 \overrightarrow{{}c}|^{2}$
$|\overrightarrow{{}a}|^{2}=\lambda^{2}(4|\overrightarrow{{}b}|^{2}+9|\overrightarrow{{}c}|^{2}+12 \overrightarrow{{}b} \cdot \overrightarrow{{}c})$
$31=31 \lambda^{2} \Rightarrow \lambda= \pm 1$
$\overrightarrow{{}a}= \pm(2 \overrightarrow{{}b}+3 \overrightarrow{{}c})$
$\frac{|\vec{a} \times \vec{c}|}{|\vec{a} \cdot \vec{b}|}=\frac{2|\vec{b} \times \vec{c}|}{2 \vec{b} \cdot \vec{b}+3 \vec{c} \cdot \vec{b}}$
$|\overrightarrow{{}b} \times \overrightarrow{{}c}|^{2}=|\overrightarrow{{}b}|^{2}|\overrightarrow{{}c}|^{2}-(\overrightarrow{{}b} \cdot \overrightarrow{{}c})^{2}=\frac{3}{4}$
$ |\frac{\vec{a} \times \vec{c}}{\vec{a} \cdot \vec{b}}| = \frac{\frac{2\cdot \sqrt{3}}{2}}{2\cdot \frac{1}{4}-\frac{3}{2}}=-\sqrt{3} $
$(\frac{\vec{a} \times \vec{c}}{\vec{a} \cdot \vec{b}})^{2} = 3$
Therefore, the answer is 3.
