Question

If \(|\overrightarrow{a}|=2,\overrightarrow{b}=2\hat{i}-\hat{j}-3\hat{k}\) and the angle between \(\overrightarrow{a}\) and \(\overrightarrow{b}\) is \(\frac{\pi}{4}\), then \(\overrightarrow{a}\cdot\overrightarrow{b}\) is equal to

• A. 4√2
• B. 2√7
• C. √30
• D. √7
• E. √14

Answer 1

The Correct Option is B

Step 1: Use the dot product formula  
The dot product of two vectors \( \overrightarrow{a} \) and \( \overrightarrow{b} \) is given by: \[ \overrightarrow{a} \cdot \overrightarrow{b} = |\overrightarrow{a}| |\overrightarrow{b}| \cos\theta \] Given: \[ |\overrightarrow{a}| = 2, \quad \overrightarrow{b} = 2\hat{i} - \hat{j} - 3\hat{k} \] The magnitude of \( \overrightarrow{b} \) is: \[ |\overrightarrow{b}| = \sqrt{2^2 + (-1)^2 + (-3)^2} \] \[ = \sqrt{4 + 1 + 9} = \sqrt{14} \] The given angle is \( \theta = \frac{\pi}{4} \).

Step 2: Compute the dot product 
Using the formula: \[ \overrightarrow{a} \cdot \overrightarrow{b} = 2 \times \sqrt{14} \times \cos\frac{\pi}{4} \] Since \( \cos\frac{\pi}{4} = \frac{\sqrt{2}}{2} \), we substitute: \[ \overrightarrow{a} \cdot \overrightarrow{b} = 2 \times \sqrt{14} \times \frac{\sqrt{2}}{2} \] \[ = \sqrt{14} \times \sqrt{2} = \sqrt{28} = 2\sqrt{7} \]

Final Answer: \( \overrightarrow{a} \cdot \overrightarrow{b} \) is \( 2\sqrt{7} \).