Vectors 1 Question 19
20. Let $\mathbf{a}$ and $\mathbf{b}$ be two unit vectors such that $\mathbf{a} \cdot \mathbf{b}=0$. For some $x, y \in R$, let $\mathbf{c}=x \mathbf{a}+y \mathbf{b}+(\mathbf{a} \times \mathbf{b})$. If $|\mathbf{c}|=2$ and the vector $\mathbf{c}$ is inclined at the same angle $\alpha$ to both $\mathbf{a}$ and $\mathbf{b}$, then the value of $8 \cos ^{2} \alpha$ is
(2018 Adv.)
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Answer:
Correct Answer: 20. (a)
Solution:
- We have,
$$ \begin{aligned} \vec{c} & =x \vec{a}+y \vec{b}+\vec{a} \times \vec{b} \text { and } \vec{a} \cdot \vec{b}=0 \\ |\vec{a}| & =|\vec{b}|=1 \text { and }|\vec{c}|=2 \end{aligned} $$
Also, given $\vec{c}$ is inclined on $\vec{a}$ and $\vec{b}$ with same angle $\alpha$
$$ \therefore \quad \vec{a} \cdot \vec{c}=x|\vec{a}|^{2}+y(\vec{a} \cdot \vec{b})+\vec{a} \cdot(\vec{a} \times \vec{b}) $$
$|\vec{a}||\vec{c}| \cos \alpha=x+0+0$
$x=2 \cos \alpha$
Similarly,
$$ \begin{aligned} & \quad|\vec{b}||\vec{c}| \cos \alpha=0+y+0 \\ & \Rightarrow y=2 \cos \alpha \\ & \qquad \begin{aligned} |\vec{c}|^{2} & =x^{2}+y^{2}+|\vec{a} \times \vec{b}|^{2} \\ 4 & =8 \cos ^{2} \alpha+|a|^{2}|b|^{2} \sin ^{2} 90^{\circ} \\ & =8 \cos ^{2} \alpha+1 \quad \Rightarrow \quad 8 \cos ^{2} \alpha=3 \end{aligned} \end{aligned} $$
