Question

Let \( \vec{a} \times \vec{b} = 7\hat{i} - 5\hat{j} - 4\hat{k} \) and \( \vec{a} = \hat{i} + 3\hat{j} - 2\hat{k} \), if the length of projection of \( \vec{b} \) on \( \vec{a} \) is \( \frac{8}{\sqrt{14}} \), then \( |\vec{b}| \) is:

• A. \( 121 \)
• B. \( \sqrt{12} \)
• C. \( \sqrt{11} \)
• D. \( 144 \)

Hint:

When dealing with vector projections and cross products, recall that the magnitude of the cross product gives the area, and the projection of one vector on another is calculated using the dot product formula.

Answer 1

The Correct Option is C

 \[ \vec{a} \times \vec{b} = 7\hat{i} - 5\hat{j} - 4\hat{k} \] \[ \vec{a} = \hat{i} + 3\hat{j} - 2\hat{k} \] The length of the projection of \( \vec{b} \) on \( \vec{a} \) is \( \frac{8}{\sqrt{14}} \). 

Step 1: Find \( |\vec{a}| \) 
\[ |\vec{a}| = \sqrt{(1)^2 + (3)^2 + (-2)^2} = \sqrt{1 + 9 + 4} = \sqrt{14} \] 

Step 2: Recall Projection Formula 
The projection of \( \vec{b} \) on \( \vec{a} \) is given by: \[ \text{Proj}_{\vec{a}} \vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|} \] Let \( \vec{a} \cdot \vec{b} = k \), so: \[ \frac{k}{\sqrt{14}} = \frac{8}{\sqrt{14}} \] From this, \[ k = 8 \] 

Step 3: Cross Product Magnitude Identity 
By the cross product identity: \[ |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta \] Where \( \sin \theta = \sqrt{1 - \cos^2 \theta} \). Since \( \cos \theta = \frac{k}{|\vec{a}| |\vec{b}|} \), we get: \[ \sin \theta = \sqrt{1 - \left(\frac{8}{\sqrt{14} |\vec{b}|}\right)^2} \] Now, \[ |\vec{a} \times \vec{b}| = \sqrt{(7)^2 + (-5)^2 + (-4)^2} = \sqrt{49 + 25 + 16} = \sqrt{90} \] \[ \sqrt{90} = \sqrt{14} |\vec{b}| \sin \theta \] \[ \sin \theta = \sqrt{1 - \left(\frac{8}{\sqrt{14} |\vec{b}|} \right)^2} = \sqrt{\frac{|\vec{b}|^2 \cdot 14 - 64}{14 |\vec{b}|^2}} \] Now, \[ \sqrt{90} = \sqrt{14} |\vec{b}| \cdot \sqrt{\frac{14|\vec{b}|^2 - 64}{14|\vec{b}|^2}} \] Equating and simplifying, \[ 90 = 14|\vec{b}|^2 - 64 \] \[ 14|\vec{b}|^2 = 154 \] \[ |\vec{b}|^2 = 11 \] \[ |\vec{b}| = \sqrt{11} \] 

Final Answer: (C) \( \sqrt{11} \)