If \( \vec{a}=\hat{\imath}-2\hat{\jmath}+3\hat{k} \) and \( \vec{b}=\hat{\imath}+2\hat{\jmath}+\hat{k} \), then which of the following statements are correct ?
(A) \( \vec{a} \) and \( \vec{b} \) are perpendicular to each other.
(B) \( \vec{a} \) and \( \vec{b} \) are parallel.
(C) \( |\vec{a}\times\vec{b}|=2\sqrt{21} \).
(D) Projection of \( \vec{a} \) on \( \vec{b}=0 \).
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A), (C) and (D) only
\( \displaystyle \vec{a}\cdot\vec{b}=1-4+3=0 \)
\( \displaystyle \vec{a}\times\vec{b}=\begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\1 & -2 & 3\\1 & 2 & 1\end{vmatrix} \)
\( \displaystyle =(-2-6)\hat{\imath}-(1-3)\hat{\jmath}+(2+2)\hat{k} \)
\( \displaystyle =-8\hat{\imath}+2\hat{\jmath}+4\hat{k} \)
\( \displaystyle |\vec{a}\times\vec{b}|=\sqrt{64+4+16} \)
\( \displaystyle =\sqrt{84}=2\sqrt{21} \)
\( \displaystyle \text{Projection of }\vec{a}\text{ on }\vec{b}=\frac{\vec{a}\cdot\vec{b}}{|\vec{b}|}=\frac{0}{\sqrt{6}}=0 \)
So \( \vec{a}\cdot\vec{b}=0 \) (both vectors non-zero), \( |\vec{a}\times\vec{b}|=2\sqrt{21} \) and the projection of \( \vec{a} \) on \( \vec{b} \) is 0.
Checking the statements
(A) \( \vec{a} \) and \( \vec{b} \) are perpendicular to each other. — Correct. The dot product is 0 and both vectors are non-zero.
(B) \( \vec{a} \) and \( \vec{b} \) are parallel. — Incorrect. The components are not proportional (1 : 1 but −2 : 2).
(C) \( |\vec{a}\times\vec{b}|=2\sqrt{21} \). — Correct. \( \sqrt{64+4+16}=\sqrt{84}=2\sqrt{21} \).
(D) Projection of \( \vec{a} \) on \( \vec{b}=0 \). — Correct. The projection is \( \frac{0}{\sqrt{6}}=0 \).
Shortcut: Since the vectors are perpendicular, the cross product needs no determinant:
\( \displaystyle |\vec{a}\times\vec{b}|=|\vec{a}|\,|\vec{b}|\sin 90^\circ \)
\( \displaystyle =\sqrt{14}\times\sqrt{6} \)
\( \displaystyle =2\sqrt{21} \)