If \( \vec{a}=\hat{\imath}-2\hat{\jmath}+3\hat{k} \) and \( \vec{b}=2\hat{\imath}+3\hat{\jmath}-5\hat{k} \), then which of the following statements are TRUE?
(A) both \( \vec{a} \) and \( \vec{b} \) are unit vectors
(B) \( \vec{a}.\vec{b}=-19 \)
(C) \( |\vec{a}\times\vec{b}|=170 \)
(D) \( \vec{a}\times\vec{b} \) is perpendicular to \( \vec{b} \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (B) and (D) only
The magnitudes and the dot product:
\( \displaystyle |\vec{a}|=\sqrt{1+4+9}=\sqrt{14} \)
\( \displaystyle |\vec{b}|=\sqrt{4+9+25}=\sqrt{38} \)
\( \displaystyle \vec{a}\cdot\vec{b}=2-6-15=-19 \)
The cross product:
\( \displaystyle \vec{a}\times\vec{b}=\begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\1 & -2 & 3\\2 & 3 & -5\end{vmatrix} \)
\( \displaystyle =\hat{\imath}(10-9)-\hat{\jmath}(-5-6)+\hat{k}(3+4) \)
\( \displaystyle =\hat{\imath}+11\hat{\jmath}+7\hat{k} \)
\( \displaystyle |\vec{a}\times\vec{b}|=\sqrt{1+121+49}=\sqrt{171} \)
\( \displaystyle (\vec{a}\times\vec{b})\cdot\vec{b}=2+33-35=0 \)
So \( |\vec{a}|=\sqrt{14} \), \( |\vec{b}|=\sqrt{38} \), \( \vec{a}\cdot\vec{b}=-19 \), \( |\vec{a}\times\vec{b}|=\sqrt{171} \), and \( \vec{a}\times\vec{b} \) is perpendicular to \( \vec{b} \).
Checking the statements
(A) both \( \vec{a} \) and \( \vec{b} \) are unit vectors — Incorrect. The magnitudes are \( \sqrt{14} \) and \( \sqrt{38} \), not 1.
(B) \( \vec{a}.\vec{b}=-19 \) — Correct. \( 2-6-15=-19 \).
(C) \( |\vec{a}\times\vec{b}|=170 \) — Incorrect. \( |\vec{a}\times\vec{b}|=\sqrt{171} \).
(D) \( \vec{a}\times\vec{b} \) is perpendicular to \( \vec{b} \) — Correct. The cross product is perpendicular to both vectors; its dot product with \( \vec{b} \) is 0.
Shortcut: Lagrange’s identity gives \( |\vec{a}\times\vec{b}| \) without the cross product:
\( \displaystyle |\vec{a}\times\vec{b}|^2=|\vec{a}|^2|\vec{b}|^2-(\vec{a}\cdot\vec{b})^2 \)
\( \displaystyle =14\times 38-361=171 \)