Let \(\vec a =\hat{\imath} +4\hat{\jmath} +2\hat{k}, \vec b=3\hat{\imath}-2\hat{\jmath}+7\hat{k}\) and \(\vec c=2\hat{\imath}+\hat{\jmath}+4\hat{k}\). A vector \(\vec d\) which is perpendicular to both \(\vec a\) and \(\vec b\) and \(\vec c.\vec d=14\), is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(64\hat{\imath}-2\hat{\jmath}-28\hat{k}\)
A vector perpendicular to both \(\vec a\) and \(\vec b\) is parallel to the cross product \(\vec a \times \vec b\):
\(\displaystyle \vec a \times \vec b = \begin{vmatrix}\hat{\imath}&\hat{\jmath}&\hat{k}\\1&4&2\\3&-2&7\end{vmatrix}\)
\(= (28 + 4)\hat{\imath} - (7 - 6)\hat{\jmath} + (-2 - 12)\hat{k}\)
\(= 32\hat{\imath} - \hat{\jmath} - 14\hat{k}\)
Let \(\vec d = \lambda(32\hat{\imath} - \hat{\jmath} - 14\hat{k})\). The condition \(\vec c \cdot \vec d = 14\) gives
\(\lambda(2 \times 32 + 1 \times (-1) + 4 \times (-14)) = 14\)
\(\lambda(64 - 1 - 56) = 14\)
\(7\lambda = 14,\quad \lambda = 2\)
So
\(\vec d = 64\hat{\imath} - 2\hat{\jmath} - 28\hat{k}\)
Shortcut: Take the dot product of each option with \(\vec a = (1, 4, 2)\). Only Option (2) gives 0: \(64 - 8 - 56 = 0\). Option (1) gives \(64 + 8 - 56 = 16\), Option (3) gives \(64 + 8 + 56 = 128\) and Option (4) gives \(32 + 4 + 28 = 64\).