Let \( \vec{a}=\hat{\imath}-\hat{\jmath},\ \vec{b}=3\hat{\jmath}-\hat{k} \) and \( \vec{c}=7\hat{\imath}-\hat{k} \), then a vector \( \vec{d} \) which is perpendicular to both \( \vec{a} \) and \( \vec{b} \) and \( \vec{c}.\vec{d}=1 \) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{1}{4}(\hat{\imath}+\hat{\jmath}+3\hat{k}) \)
Since \( \vec{d} \) is perpendicular to both \( \vec{a} \) and \( \vec{b} \), it is along \( \vec{a}\times\vec{b} \).
\( \displaystyle \vec{a}\times\vec{b}=\begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\1 & -1 & 0\\0 & 3 & -1\end{vmatrix} \)
\( \displaystyle =(1-0)\hat{\imath}-(-1-0)\hat{\jmath}+(3-0)\hat{k} \)
\( \displaystyle =\hat{\imath}+\hat{\jmath}+3\hat{k} \)
Let \( \vec{d}=t(\hat{\imath}+\hat{\jmath}+3\hat{k}) \). Then
\( \vec{c}\cdot\vec{d}=t(7+0-3)=4t \)
\( \displaystyle 4t=1\;\Rightarrow\;t=\frac{1}{4} \)
Why the other options are incorrect
Option (1) \( (\hat{\imath}+\hat{\jmath}+3\hat{k}) \) — Incorrect. This vector is perpendicular to both, but gives \( \vec{c}\cdot\vec{d}=4 \), not 1.
Option (3) \( \displaystyle\frac{1}{4}(\hat{\imath}+\hat{\jmath}-3\hat{k}) \) — Incorrect. Its dot product with \( \vec{b} \) is \( \frac{1}{4}(3+3)\ne 0 \), so it is not perpendicular to \( \vec{b} \).
Option (4) \( \displaystyle\frac{1}{4}(\hat{\imath}-\hat{\jmath}+3\hat{k}) \) — Incorrect. Its dot product with \( \vec{a} \) is \( \frac{1}{4}(1+1)\ne 0 \), so it is not perpendicular to \( \vec{a} \).