Question

Let \(\vec{a} = 4\hat{i} - \hat{j} + \hat{k}\), \(\vec{b} = 11\hat{i} - \hat{j} + \hat{k}\), and \(\vec{c}\) be a vector such that \[ (\vec{a} + \vec{b}) \times \vec{c} = \vec{c} \times (-2\vec{a} + 3\vec{b}). \] If \((2\vec{a} + 3\vec{b}) \cdot \vec{c} = 1670\), then \(|\vec{c}|^2\) is equal to:

• A. 1627
• B. 1618
• C. 1600
• D. 1609

Answer 1

The Correct Option is B

Given: \[ (\vec{a} + \vec{b}) \times \vec{c} = \vec{c} \times (-2\vec{a} + 3\vec{b}). \]

Expanding: \[ (\vec{a} + \vec{b}) \times \vec{c} + \vec{c} \times (-2\vec{a} + 3\vec{b}) = 0. \]

This implies that: \[ \vec{c} = \lambda (4\vec{b} - \vec{a}). \] 

We substitute the values of \(\vec{a}\) and \(\vec{b}\): \[ \vec{c} = \lambda \left( 4(11\hat{i} - \hat{j} + \hat{k}) - (4\hat{i} - \hat{j} + \hat{k}) \right), \] \[ \vec{c} = \lambda (40\hat{i} - 3\hat{j} + 3\hat{k}). \] 

Given that: \[ (2\vec{a} + 3\vec{b}) \cdot \vec{c} = 1670. \] 

Calculating \( 2\vec{a} + 3\vec{b} \): \[ 2\vec{a} + 3\vec{b} = 2(4\hat{i} - \hat{j} + \hat{k}) + 3(11\hat{i} - \hat{j} + \hat{k}), \] \[ 2\vec{a} + 3\vec{b} = (8 + 33)\hat{i} + (-2 - 3)\hat{j} + (2 + 3)\hat{k}, \] \[ 2\vec{a} + 3\vec{b} = 41\hat{i} - 5\hat{j} + 5\hat{k}. \] 

Now, we compute: \[ (2\vec{a} + 3\vec{b}) \cdot \vec{c} = (41\hat{i} - 5\hat{j} + 5\hat{k}) \cdot \lambda (40\hat{i} - 3\hat{j} + 3\hat{k}), \] \[ (2\vec{a} + 3\vec{b}) \cdot \vec{c} = \lambda (41 \cdot 40 + (-5) \cdot (-3) + 5 \cdot 3), \] \[ 1670 = \lambda (1640 + 15 + 15), \] \[ 1670 = 1670\lambda \implies \lambda = 1. \] 

Thus: \[ \vec{c} = 40\hat{i} - 3\hat{j} + 3\hat{k}. \] 

Calculating \( |\vec{c}|^2 \): \[ |\vec{c}|^2 = 40^2 + (-3)^2 + 3^2, \] \[ |\vec{c}|^2 = 1600 + 9 + 9, \] \[ |\vec{c}|^2 = 1618. \] 

Therefore: \[ 1618. \]