For the vector \( \vec{a}=\displaystyle\frac{1}{3}\left(2\hat{\imath}-2\hat{\jmath}+\hat{k}\right) \), which of the following statements is/are TRUE ?
A. vector \( \vec{a} \) is a unit vector
B. vector \( \vec{a} \) is making an angle \( \displaystyle\frac{\pi}{3} \) with vector \( 2\hat{\imath}-4\hat{\jmath}+3\hat{k} \)
C. vector \( \vec{a} \) is parallel to vector \( -\hat{\imath}+\hat{\jmath}-\displaystyle\frac{1}{2}\hat{k} \)
D. vector \( \vec{a} \) is perpendicular to vector \( 3\hat{\imath}+2\hat{\jmath}-2\hat{k} \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → A, C and D only
\( \displaystyle |\vec{a}|=\frac{1}{3}\sqrt{4+4+1} \)
\( \displaystyle =1 \)
For the angle with \( 2\hat{\imath}-4\hat{\jmath}+3\hat{k} \), whose length is \( \sqrt{29} \):
\( \displaystyle \cos\theta=\frac{\frac{1}{3}(4+8+3)}{1\times\sqrt{29}} \)
\( \displaystyle =\frac{5}{\sqrt{29}} \)
\( \displaystyle -\hat{\imath}+\hat{\jmath}-\frac{1}{2}\hat{k}=-\frac{1}{2}(2\hat{\imath}-2\hat{\jmath}+\hat{k})=-\frac{3}{2}\,\vec{a} \)
\( \displaystyle \vec{a}\cdot(3\hat{\imath}+2\hat{\jmath}-2\hat{k})=\frac{1}{3}(6-4-2) \)
\( \displaystyle =0 \)
So \( |\vec{a}|=1 \), \( \cos\theta=\frac{5}{\sqrt{29}} \), \( -\hat{\imath}+\hat{\jmath}-\frac{1}{2}\hat{k} \) is a scalar multiple of \( \vec{a} \), and \( \vec{a}\cdot(3\hat{\imath}+2\hat{\jmath}-2\hat{k})=0 \).
Checking the statements
(A) vector \( \vec{a} \) is a unit vector — Correct. \( |\vec{a}|=1 \).
(B) vector \( \vec{a} \) is making an angle \( \displaystyle\frac{\pi}{3} \) with vector \( 2\hat{\imath}-4\hat{\jmath}+3\hat{k} \) — Incorrect. \( \cos\theta=\frac{5}{\sqrt{29}} \), not \( \frac{1}{2} \).
(C) vector \( \vec{a} \) is parallel to vector \( -\hat{\imath}+\hat{\jmath}-\displaystyle\frac{1}{2}\hat{k} \) — Correct. This vector is \( -\frac{3}{2}\,\vec{a} \).
(D) vector \( \vec{a} \) is perpendicular to vector \( 3\hat{\imath}+2\hat{\jmath}-2\hat{k} \) — Correct. The dot product is \( \frac{1}{3}(6-4-2)=0 \).