Let overrightarrow{a}=2 hat{i}-hat{j}+hat{k} | vedclass

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(C) Given $\overrightarrow{a}=2 \hat{i}-\hat{j}+\hat{k}$ and $\overrightarrow{b}=\lambda \hat{j}+2 \hat{k}$.$\overrightarrow{c}=\overrightarrow{a} 	i

Vedclass · Accepted Answer

$208$

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(C) Given $\overrightarrow{a}=2 \hat{i}-\hat{j}+\hat{k}$ and $\overrightarrow{b}=\lambda \hat{j}+2 \hat{k}$.$\overrightarrow{c}=\overrightarrow{a} 	imes \overrightarrow{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -1 & 1 \ 0 & \lambda & 2 \end{vmatrix} = (-2-\lambda) \hat{i} - 4 \hat{j} + 2 \lambda \hat{k}$.Given $|\overrightarrow{c}|=\sqrt{53}$,so $|\overrightarrow{c}|^2 = 53$.$(-2-\lambda)^2 + (-4)^2 + (2\lambda)^2 = 53$$4 + 4\lambda + \lambda^2 + 16 + 4\lambda^2 = 53$$5\lambda^2 + 4\lambda - 33 = 0$.Solving for $\lambda$: $\lambda = \frac{-4 \pm \sqrt{16 - 4(5)(-33)}}{10} = \frac{-4 \pm \sqrt{16 + 660}}{10} = \frac{-4 \pm \sqrt{676}}{10} = \frac{-4 \pm 26}{10}$.Since $\lambda \in \mathbb{Z}$,we take $\lambda = -3$ (as $\frac{22}{10}$ is not an integer).Thus,$\overrightarrow{c} = (-2 - (-3))\hat{i} - 4\hat{j} + 2(-3)\hat{k} = \hat{i} - 4\hat{j} - 6\hat{k}$.Let $\overrightarrow{d} = y\hat{j} + z\hat{k}$ be a vector in the $yz$-plane with $|\overrightarrow{d}|=2$,so $y^2 + z^2 = 4$.Then $\overrightarrow{c} \cdot \overrightarrow{d} = (\hat{i} - 4\hat{j} - 6\hat{k}) \cdot (y\hat{j} + z\hat{k}) = -4y - 6z$.We want to maximize $(-4y - 6z)^2 = (4y + 6z)^2$.By Cauchy-Schwarz inequality,$(4y + 6z)^2 \leq (4^2 + 6^2)(y^2 + z^2) = (16 + 36)(4) = 52 	imes 4 = 208$.

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