Vectors 2 Question 5
5. Let $\mathbf{a}=3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+x \hat{\mathbf{k}}$ and $\mathbf{b}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$, for some real $x$. Then $|\mathbf{a} \times \mathbf{b}|=r$ is possible if
(2019 Main, 8 April II)
(a) $0<r \leq \sqrt{\frac{3}{2}}$
(b) $\sqrt{\frac{3}{2}}<r \leq 3 \sqrt{\frac{3}{2}}$
(c) $3 \sqrt{\frac{3}{2}}<r<5 \sqrt{\frac{3}{2}}$
(d) $r \geq 5 \sqrt{\frac{3}{2}}$
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Answer:
Correct Answer: 5. (d)
Solution:
- Given vectors are $\mathbf{a}=3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+x \hat{\mathbf{k}}$
and $\mathbf{b}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$
$$ \begin{aligned} \therefore \mathbf{a} \times \mathbf{b}= & \left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 3 & 2 & x \\ 1 & -1 & 1 \end{array}\right|=\hat{\mathbf{i}}(2+x)-\hat{\mathbf{j}}(3-x)+\hat{\mathbf{k}}(-3-2) \\ & =(x+2) \hat{\mathbf{i}}+(x-3) \hat{\mathbf{j}}-5 \hat{\mathbf{k}} \\ \Rightarrow|\mathbf{a} \times \mathbf{b}| & =\sqrt{(x+2)^{2}+(x-3)^{2}+25} \\ & =\sqrt{2 x^{2}-2 x+4+9+25} \\ & =\sqrt{2\left(x^{2}-x+\frac{1}{4}\right)-\frac{1}{2}+38}=\sqrt{2\left(x-\frac{1}{2}\right)^{2}+\frac{75}{2}} \end{aligned} $$
$$ =\sqrt{2\left(x^{2}-x+\frac{1}{4}\right)-\frac{1}{2}+38}=\sqrt{2\left(x-\frac{1}{2}\right)^{2}+\frac{75}{2}} $$
So, $|\mathbf{a} \times \mathbf{b}| \geq \sqrt{\frac{75}{2}} \quad$ [at $x=\frac{1}{2},|\mathbf{a} \times \mathbf{b}|$ is minimum]
$\Rightarrow \quad r \geq 5 \sqrt{\frac{3}{2}}$
