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Question

Let $\overrightarrow{a} = 2\widehat{i} - \widehat{j} + \widehat{k}$ and $\overrightarrow{b} = \lambda\widehat{j} + 2\widehat{k},\lambda \in Z$ be two vectors, Let $\overrightarrow{c} = \overrightarrow{a} \times \overrightarrow{b}$ and $\overrightarrow{d}$ be a vector of magnitude 2 in yz-plane. If $|\overrightarrow{c}| = \sqrt{53}$, then the maximum possible value of $(\overrightarrow{c} \cdot \overrightarrow{d})^{2}$ is equal to :

Options

A.26
B.104
C.208
D.52

Solution

$\overrightarrow{a} = 2\widehat{i} - \widehat{j} + \widehat{k}$

$${\overrightarrow{b} = \lambda\widehat{j} + 2\widehat{k};\lambda \in Z }{\overrightarrow{c} = \overrightarrow{a} \times \overrightarrow{b} = ( - 2 - \lambda)\widehat{i} - 4\widehat{j} + 2\lambda\widehat{k} }{|\overrightarrow{c}| = \sqrt{53} }{\Rightarrow 5\lambda^{2} + 4\lambda - 33 = 0 }$$$\lambda = 2.2$ or -3

$${\Rightarrow \lambda = - 3 }{\overrightarrow{c} = \widehat{i} - 4\widehat{j} - 6\widehat{k} }$$let $\overrightarrow{d} = y\widehat{j} + z\widehat{k}$

$${|\overrightarrow{d}| = 2 }{\Rightarrow y^{2} + z^{2} = 4 }{(\overrightarrow{c} \cdot \overrightarrow{d})^{2} = (4y + 6z)^{2} \leq \left( \sqrt{4^{2} + 6^{2}} \times \sqrt{y^{2} + z^{2}} \right)^{2} \leq 208}$$

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