If $\vec a = 2\hat i-3\hat j+\hat k$ and $\vec b = 2\hat i+\hat j-\hat k$, then which of the following statements is/are correct?
(A) $\vec a$ and $\vec b$ are collinear
(B) $\vec a$ and $\vec b$ are perpendicular
(C) Angle between $\vec a$ and $\vec b$ is $\frac{\pi}{4}$
(D) $|\vec a +\vec b| = 2\sqrt{5}$
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (B) and (D) only
Given:
$\vec a = 2\hat{i} - 3\hat{j} + \hat{k}$
$\vec b = 2\hat{i} + \hat{j} - \hat{k}$
1. Check collinearity:
If $\vec a$ and $\vec b$ are collinear, then $\vec b = \lambda \vec a$.
But the ratios:
$\frac{2}{2} = 1$, $\frac{1}{-3} = -\frac{1}{3}$, $\frac{-1}{1} = -1$
They are not equal → Not collinear.
2. Check perpendicularity:
$\vec a \cdot \vec b = (2)(2) + (-3)(1) + (1)(-1)$
$= 4 - 3 - 1 = 0$
Dot product = 0 → Vectors are perpendicular.
3. Angle between $\vec a$ and $\vec b$:
Since $\vec a \cdot \vec b = 0$, angle = $\frac{\pi}{2}$, not $\frac{\pi}{4}$.
4. Magnitude of $\vec a + \vec b$:
$\vec a + \vec b = (2+2)\hat i + (-3+1)\hat j + (1 - 1)\hat k$
$= 4\hat i - 2\hat j$
Magnitude:
$|\vec a + \vec b| = \sqrt{4^2 + (-2)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}$
Final Answer: (B) and (D)