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JEE Mains · Maths · STD 12 - 10. vector algebra

Let \(\vec{a}=2 \hat{i}+3 \hat{j}+4 \hat{k}, \vec{b}=2 \hat{i}-2 \hat{j}-2 \hat{k}\) and \(\overrightarrow{ c }=-\hat{ i }+4 \hat{ j }+3 \hat{ k }\). If \(\overrightarrow{ d }\) is a vector perpendicular to both \(\vec{b}\) and \(\overrightarrow{ c }\) and \(\overrightarrow{ a } \cdot \overrightarrow{ d }=18\), Then \(|\overrightarrow{ a } \times \overrightarrow{ d }|^2\) is equal to \(..........\).

  1. A \(640\)
  2. B \(760\)
  3. C \(680\)
  4. D \(720\)
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Answer & Solution

Correct Answer

(D) \(720\)

Step-by-step Solution

Detailed explanation

\(\overrightarrow{ a }=\lambda(\overrightarrow{ b } \times \overrightarrow{ c })\) \(\overrightarrow{ b } \times \overrightarrow{ c }=\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \\ 1 & -2 & -2 \\ -1 & 4 & 3\end{array}\right|=2 \hat{i}-\hat{ j }+2 \hat{ k }\)…
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