For the vectors \( \vec{a}=\hat{\imath}-2\hat{\jmath}+3\hat{k} \) and \( \vec{b}=3\hat{\imath}-2\hat{\jmath}+\hat{k} \). Which of the following are correct ?
(A) \( |\vec{a}|=14 \)
(B) \( |\vec{a}|=|\vec{b}| \)
(C) \( \vec{a}.\vec{b}=10 \)
(D) angle between \( \vec{a} \) and \( \vec{b} \) is \( \cos^{-1}\left(\displaystyle\frac{5}{7}\right) \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (B), (C) and (D) only
\( \displaystyle |\vec{a}|=\sqrt{1+4+9}=\sqrt{14} \)
\( \displaystyle |\vec{b}|=\sqrt{9+4+1}=\sqrt{14} \)
\( \displaystyle \vec{a}\cdot\vec{b}=(1)(3)+(-2)(-2)+(3)(1) \)
\( \displaystyle =3+4+3=10 \)
\( \displaystyle \cos\theta=\frac{\vec{a}\cdot\vec{b}}{|\vec{a}|\,|\vec{b}|} \)
\( \displaystyle =\frac{10}{\sqrt{14}\sqrt{14}} \)
\( \displaystyle =\frac{10}{14} \)
\( \displaystyle =\frac{5}{7} \)
So \( |\vec{a}|=|\vec{b}|=\sqrt{14} \), \( \vec{a}\cdot\vec{b}=10 \) and the angle between them is \( \cos^{-1}\left(\frac{5}{7}\right) \).
Checking the statements
(A) \( |\vec{a}|=14 \) — Incorrect. \( |\vec{a}|=\sqrt{14} \); 14 is \( |\vec{a}|^2 \).
(B) \( |\vec{a}|=|\vec{b}| \) — Correct. Both are \( \sqrt{14} \).
(C) \( \vec{a}.\vec{b}=10 \) — Correct. 3 + 4 + 3 = 10.
(D) angle between \( \vec{a} \) and \( \vec{b} \) is \( \cos^{-1}\left(\frac{5}{7}\right) \) — Correct. \( \cos\theta=\frac{10}{14}=\frac{5}{7} \).