Let overrightarrow{x} be a vector in the plan | vedclass

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Question

(C) Since $\overrightarrow{x}$ lies in the plane of $\overrightarrow{a}$ and $\overrightarrow{b}$,we can write $\overrightarrow{x} = \lambda\overright

Vedclass · Accepted Answer

$486$

Vedclass · Answer

(C) Since $\overrightarrow{x}$ lies in the plane of $\overrightarrow{a}$ and $\overrightarrow{b}$,we can write $\overrightarrow{x} = \lambda\overrightarrow{a} + \mu\overrightarrow{b} = \lambda(2\hat{i} - \hat{j} + \hat{k}) + \mu(\hat{i} + 2\hat{j} - \hat{k}) = (2\lambda + \mu)\hat{i} + (2\mu - \lambda)\hat{j} + (\lambda - \mu)\hat{k}$.Given $\overrightarrow{x} \perp (3\hat{i} + 2\hat{j} - \hat{k})$,so $\overrightarrow{x} \cdot (3\hat{i} + 2\hat{j} - \hat{k}) = 0$.$3(2\lambda + \mu) + 2(2\mu - \lambda) - 1(\lambda - \mu) = 0 \implies 6\lambda + 3\mu + 4\mu - 2\lambda - \lambda + \mu = 0 \implies 3\lambda + 8\mu = 0 \implies \lambda = -\frac{8}{3}\mu$.Projection of $\overrightarrow{x}$ on $\overrightarrow{a}$ is $\frac{\overrightarrow{x} \cdot \overrightarrow{a}}{|\overrightarrow{a}|} = \frac{17\sqrt{6}}{2}$.$|\overrightarrow{a}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{6}$.$\overrightarrow{x} \cdot \overrightarrow{a} = (2\lambda + \mu)(2) + (2\mu - \lambda)(-1) + (\lambda - \mu)(1) = 4\lambda + 2\mu - 2\mu + \lambda + \lambda - \mu = 6\lambda - \mu$.So,$\frac{6\lambda - \mu}{\sqrt{6}} = \frac{17\sqrt{6}}{2} \implies 6\lambda - \mu = 51$.Substituting $\lambda = -\frac{8}{3}\mu$ into $6\lambda - \mu = 51$: $6(-\frac{8}{3}\mu) - \mu = 51 \implies -16\mu - \mu = 51 \implies -17\mu = 51 \implies \mu = -3$.Then $\lambda = -\frac{8}{3}(-3) = 8$.Thus,$\overrightarrow{x} = 8(2\hat{i} - \hat{j} + \hat{k}) - 3(\hat{i} + 2\hat{j} - \hat{k}) = (16-3)\hat{i} + (-8-6)\hat{j} + (8+3)\hat{k} = 13\hat{i} - 14\hat{j} + 11\hat{k}$.$|\overrightarrow{x}|^{2} = 13^2 + (-14)^2 + 11^2 = 169 + 196 + 121 = 486$.

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