Question

Let \(\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}\), \(\vec{b} = 2\hat{i} + 3\hat{j} - 5\hat{k}\), and \(\vec{c} = 3\hat{i} - \hat{j} + \lambda\hat{k}\) be three vectors. Let \(\vec{r}\) be a unit vector along \(\vec{b} + \vec{c}\). If \(\vec{r} \cdot \vec{a} = 3\), then \(3\lambda\) is equal to:

• A. 27
• B. 25
• C. 25
• D. 21

Answer 1

The Correct Option is B

Given vectors:

\[ \vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}, \quad \vec{b} = 2\hat{i} + 3\hat{j} - 5\hat{k}, \quad \vec{c} = 3\hat{i} - \hat{j} + \lambda \hat{k}. \] 

The sum of vectors \(\vec{b} + \vec{c}\) is given by: 

\[ \vec{b} + \vec{c} = (2\hat{i} + 3\hat{j} - 5\hat{k}) + (3\hat{i} - \hat{j} + \lambda \hat{k}) = (5\hat{i} + 2\hat{j} + (\lambda - 5)\hat{k}). \] The magnitude of \(\vec{b} + \vec{c}\) is: \[ |\vec{b} + \vec{c}| = \sqrt{5^2 + 2^2 + (\lambda - 5)^2} = \sqrt{25 + 4 + (\lambda - 5)^2}. \] 

Simplifying:

 \[ |\vec{b} + \vec{c}| = \sqrt{29 + (\lambda - 5)^2}. \] 

The unit vector along \(\vec{b} + \vec{c}\) is:

 \[ \vec{r} = \frac{\vec{b} + \vec{c}}{|\vec{b} + \vec{c}|} = \frac{5\hat{i} + 2\hat{j} + (\lambda - 5)\hat{k}}{\sqrt{29 + (\lambda - 5)^2}}. \] Given that \(\vec{r} \cdot \vec{a} = 3\), 

we have: \[ \frac{1}{\sqrt{29 + (\lambda - 5)^2}} (5 \cdot 1 + 2 \cdot 2 + 3 \cdot (\lambda - 5)) = 3. \] 

Simplifying: 
\[ \frac{1}{\sqrt{29 + (\lambda - 5)^2}} (5 + 4 + 3\lambda - 15) = 3. \] 
\[ \frac{3\lambda - 6}{\sqrt{29 + (\lambda - 5)^2}} = 3. \] 

Cross-multiplying: 
\[ \lambda - 2 = \sqrt{29 + (\lambda - 5)^2}. \] 

Squaring both sides: 
\[ (\lambda - 2)^2 = 29 + (\lambda - 5)^2. \] 

Expanding both sides: 
\[ \lambda^2 - 4\lambda + 4 = 29 + \lambda^2 - 10\lambda + 25. \] Simplifying: 
\[ -4\lambda + 4 = 54 - 10\lambda. \] 

Rearranging terms: 
\[ 6\lambda = 50 \implies \lambda = \frac{50}{6} = \frac{25}{3}. \] 

Thus: 
\[ 3\lambda = 3 \times \frac{25}{3} = 25. \] Therefore: 
\[ 25. \]