Let a = 2 i - j + k and b = j + 2 k be two vectors. Let c = a × b a n d d be a vector of magnitude 2 in yz-plane. If | c | = 53 , then the maximum possible value of ( c · d) 2 is equal to: [2026]

Question

Let $\vec{a} = 2 \hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = λ \hat{j} + 2 \hat{k}, λ \in Z$ be two vectors. Let $\vec{c} = \vec{a} \times \vec{b} and \vec{d}$ be a vector of magnitude 2 in yz-plane. If $|| \vec{c} || = \sqrt{53},$ then the maximum possible value of ${(\vec{c} \cdot \vec{d})}^{2}$ is equal to: [2026]

Smart Achievers · Accepted Answer

(4)

$\vec{a} = 2 \hat{i} - \hat{j} + \hat{k}$

$\vec{b} = λ \hat{j} + 2 \hat{k}, λ \in ℤ$

$\vec{c} = \vec{a} \times \vec{b} = (- 2 - λ) \hat{i} - 4 \hat{j} + 2 λ \hat{k}$

$| \vec{c} |$ $= \sqrt{53}$

$\Rightarrow 5 λ^{2} + 4 λ - 33 = 0$

$λ = 2.2 or - 3$

$\Rightarrow λ = - 3$

$\vec{c} = \hat{i} - 4 \hat{j} - 6 \hat{k}$

$let \vec{d} = y \hat{j} + z \hat{k}$

$| \vec{d} |$ $= 2$

$\Rightarrow y^{2} + z^{2} = 4$

${(\vec{c} \cdot \vec{d})}^{2} = {(4 y + 6 z)}^{2}$ $\leq {(\sqrt{4^{2} + 6^{2}} \times \sqrt{y^{2} + z^{2}})}^{2} \leq 208$

Solution

1 104

2 52

3 26

4 208

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