Let \( \vec{a}=\left(\hat{\imath}+4\hat{\jmath}+2\hat{k}\right),\vec{b}=\left(3\hat{\imath}-2\hat{\jmath}+7\hat{k}\right),\vec{c}=\left(2\hat{\imath}-\hat{\jmath}+4\hat{k}\right) \). A vector \( \vec{d} \), which is perpendicular to both \( \vec{a} \) and \( \vec{b} \), such that \( \vec{c}.\vec{d}=18 \), 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 \( \vec{a}\times\vec{b} \), so \( \vec{d}=t(\vec{a}\times\vec{b}) \) for some scalar t, and the condition \( \vec{c}\cdot\vec{d}=18 \) fixes t.
\( \displaystyle \vec{a}\times\vec{b}=\begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\1 & 4 & 2\\3 & -2 & 7\end{vmatrix} \)
\( \displaystyle =\hat{\imath}(28+4)-\hat{\jmath}(7-6)+\hat{k}(-2-12) \)
\( \displaystyle =32\hat{\imath}-\hat{\jmath}-14\hat{k} \)
\( \displaystyle \vec{d}=t\left(32\hat{\imath}-\hat{\jmath}-14\hat{k}\right) \)
\( \displaystyle \vec{c}\cdot\vec{d}=t\left[(2)(32)+(-1)(-1)+(4)(-14)\right] \)
\( \displaystyle =t(64+1-56) \)
\( \displaystyle =9t \)
\( 9t=18 \)
\( t=2 \)
\( \displaystyle \vec{d}=64\hat{\imath}-2\hat{\jmath}-28\hat{k} \)
Shortcut: Check the options with dot products: the correct \( \vec{d} \) must give \( \vec{a}\cdot\vec{d}=0 \) and \( \vec{c}\cdot\vec{d}=18 \). For Option (2):
\( \displaystyle \vec{a}\cdot\vec{d}=64-8-56=0 \)
\( \displaystyle \vec{c}\cdot\vec{d}=128+2-112=18 \)
Why the other options are incorrect
Option (1) \( 32\hat{\imath}-\hat{\jmath}-14\hat{k} \) — Incorrect. This is \( \vec{a}\times\vec{b} \) itself (t = 1); it is perpendicular to both, but \( \vec{c}\cdot\vec{d}=9 \), not 18.
Option (3) \( 64\hat{\imath}+\hat{\jmath}+56\hat{k} \) — Incorrect. \( \vec{a}\cdot\vec{d}=64+4+112=180\ne 0 \), so it is not perpendicular to \( \vec{a} \).
Option (4) \( 32\hat{\imath}+\hat{\jmath}+14\hat{k} \) — Incorrect. \( \vec{a}\cdot\vec{d}=32+4+28=64\ne 0 \), so it is not perpendicular to \( \vec{a} \).