Question

(e) The angle between the vectors \( 3\hat{i} - 2\hat{j} + \hat{k} \) and \( 2\hat{i} + \hat{j} + 3\hat{k} \) will be:

• A. \( 60^\circ \)
• B. \( 30^\circ \)
• C. \( 90^\circ \)
• D. \( \cos^{-1}\left(\frac{1}{14}\right) \)

Hint:

To find the angle between two vectors, calculate the dot product and magnitudes carefully.

Answer 1

The Correct Option is D

The cosine of the angle between two vectors is given by: \[ \cos\theta=\frac{\vec{A}\cdot\vec{B}}{\|\vec{A}\|\|\vec{B}\|}. \] \[ \vec{A}=3\hat{i}-2\hat{j}+\hat{k},\quad\vec{B}=2\hat{i}+\hat{j}+3\hat{k}. \] Compute \( \vec{A}\cdot\vec{B} \): \[ \vec{A}\cdot\vec{B}=(3)(2)+(-2)(1)+(1)(3)=6-2+3=7. \] Compute \( \|\vec{A}\| \) and \( \|\vec{B}\| \): \[ \|\vec{A}\|=\sqrt{3^2+(-2)^2+1^2}=\sqrt{9+4+1}=\sqrt{14}. \] \[ \|\vec{B}\|=\sqrt{2^2+1^2+3^2}=\sqrt{4+1+9}=\sqrt{14}. \] \[ \cos\theta=\frac{7}{\sqrt{14}\cdot\sqrt{14}}=\frac{7}{14}=\frac{1}{2}. \] \[ \theta=\cos^{-1}\left(\frac{1}{2}\right). \]