Let \(\vec{a} = 2\hat{\imath} - \hat{\jmath}, \vec{b} = -4\hat{\jmath} + \hat{k}\) and \(\vec{c} = \hat{\imath} + 2\hat{k}\). If \(\vec{d}\) is a vector perpendicular to both \(\vec{a}\) and \(\vec{b}\) such that \(\vec{c} \cdot \vec{d} = 34\), then \(|\vec{d}|\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(2\sqrt{69}\)
A vector perpendicular to both \(\vec{a}\) and \(\vec{b}\) is parallel to their cross product:
\( \displaystyle \vec{a} \times \vec{b} = \begin{vmatrix}\hat{\imath}&\hat{\jmath}&\hat{k}\\2&-1&0\\0&-4&1\end{vmatrix} \)
\(= \hat{\imath}(-1 - 0) - \hat{\jmath}(2 - 0) + \hat{k}(-8 - 0)\)
\(= -\hat{\imath} - 2\hat{\jmath} - 8\hat{k}\)
So \(\vec{d} = \lambda(-\hat{\imath} - 2\hat{\jmath} - 8\hat{k})\) for some scalar \(\lambda\). Using \(\vec{c} \cdot \vec{d} = 34\):
\(\lambda\big[(1)(-1) + (0)(-2) + (2)(-8)\big] = 34\)
\(-17\lambda = 34,\quad \lambda = -2\)
\(\vec{d} = 2\hat{\imath} + 4\hat{\jmath} + 16\hat{k}\)
\(|\vec{d}| = \sqrt{4 + 16 + 256} = \sqrt{276}\)
\(= \sqrt{4 \times 69} = 2\sqrt{69}\)