Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}$, $\vec{b}=3\hat{i}+2\hat{j}-\hat{k}$, $\vec{c}=\lambda\hat{j}+\mu\hat{k}$ and $\hat{d}$ be a unit vector such that $\vec{a}\times\hat{d}=\vec{b}\times\hat{d}$ and $\vec{c}\cdot\hat{d}=1$. If $\vec{c}$ is perpendicular to $\vec{a}$, then $|3\lambda\hat{d}+\mu\vec{c}|^{2}$ is equal to
Solution:
$$\vec{a}\times\hat{d} = \vec{b}\times\hat{d}$$
$$\vec{a}\times\hat{d} – \vec{b}\times\hat{d} = 0$$
$$(\vec{a}-\vec{b})\times\hat{d} = 0$$
$$\hat{d} = t(\vec{a}-\vec{b})$$
$$\vec{a}-\vec{b} = (\hat{i}+\hat{j}+\hat{k}) – (3\hat{i}+2\hat{j}-\hat{k}) = -2\hat{i}-\hat{j}+2\hat{k}$$
$$\hat{d} = t(-2\hat{i}-\hat{j}+2\hat{k})$$
$$|\hat{d}| = 1$$
$$|t|\sqrt{(-2)^{2} + (-1)^{2} + 2^{2}} = 1$$
$$|t|\sqrt{9} = 1 \Rightarrow 3|t| = 1 \Rightarrow |t| = \frac{1}{3}$$
$$t^{2} = \frac{1}{9}$$
$$\vec{c} \cdot \vec{a} = 0$$
$$(\lambda\hat{j}+\mu\hat{k}) \cdot (\hat{i}+\hat{j}+\hat{k}) = 0$$
$$\lambda + \mu = 0 \Rightarrow \mu = -\lambda$$
$$\vec{c} = \lambda\hat{j} – \lambda\hat{k} = \lambda(\hat{j}-\hat{k})$$
$$|\vec{c}|^{2} = \lambda^{2}(1^{2} + (-1)^{2}) = 2\lambda^{2}$$
$$\vec{c} \cdot \hat{d} = 1$$
$$(\lambda\hat{j} – \lambda\hat{k}) \cdot t(-2\hat{i}-\hat{j}+2\hat{k}) = 1$$
$$\lambda(-t) + (-\lambda)(2t) = 1$$
$$-3\lambda t = 1 \Rightarrow \lambda t = -\frac{1}{3}$$
$$\lambda^{2} t^{2} = \frac{1}{9}$$
$$\lambda^{2} \left(\frac{1}{9}\right) = \frac{1}{9} \Rightarrow \lambda^{2} = 1$$
$$|3\lambda\hat{d}+\mu\vec{c}|^{2} = 9\lambda^{2}|\hat{d}|^{2} + \mu^{2}|\vec{c}|^{2} + 6\lambda\mu(\hat{d}\cdot\vec{c})$$
$$= 9\lambda^{2}(1) + (-\lambda)^{2}(2\lambda^{2}) + 6\lambda(-\lambda)(1)$$
$$= 9\lambda^{2} + 2\lambda^{4} – 6\lambda^{2}$$
$$= 3\lambda^{2} + 2\lambda^{4}$$
$$= 3(1) + 2(1)^{2} = 5$$
Ans. (5)