Let \(\vec{a} = \hat{\imath} + 4\hat{\jmath},\ \vec{b} = 4\hat{\jmath} + \hat{k}\) and \(\vec{c} = \hat{\imath} - 2\hat{k}\). If \(\vec{d}\) is a vector perpendicular to both \(\vec{a}\) and \(\vec{b}\) such that \(\vec{c}\cdot\vec{d} = 16\), then \(|\vec{d}|\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(4\sqrt{33}\)
A vector perpendicular to both \(\vec{a}\) and \(\vec{b}\) is parallel to the cross product \(\vec{a} \times \vec{b}\):
\( \displaystyle \vec{a} \times \vec{b} = \begin{vmatrix}\hat{\imath}&\hat{\jmath}&\hat{k}\\1&4&0\\0&4&1\end{vmatrix} \)
\(\vec{a} \times \vec{b} = (4 - 0)\hat{\imath} - (1 - 0)\hat{\jmath} + (4 - 0)\hat{k} = 4\hat{\imath} - \hat{\jmath} + 4\hat{k}\)
So \(\vec{d} = t(4\hat{\imath} - \hat{\jmath} + 4\hat{k})\) for some scalar t. Using \(\vec{c}\cdot\vec{d} = 16\):
\(t(1 \cdot 4 + 0 \cdot (-1) + (-2) \cdot 4) = 16\)
\(-4t = 16\)
\(t = -4\)
Hence:
\( \displaystyle |\vec{d}| = |t|\sqrt{4^2 + (-1)^2 + 4^2} = 4\sqrt{33} \)