Question

Find the value of 'b' such that the scalar product of the vector $ \hat{i} + \hat{j} + \hat{k} $ with the unit vector parallel to the sum of the vectors $ 2\hat{i} + 4\hat{j} - 5\hat{k} $ and $ b\hat{i} + 2\hat{j} + 3\hat{k} $ is unity.

• A. \( -2 \)
• B. \( 0 \)
• C. \( -1 \)
• D. \( 1 \)

Hint:

To find the value of \( b \) such that the scalar product is unity, equate the scalar product expression to 1 and solve for \( b \).

Answer 1

The Correct Option is D

Let the two vectors be \( \mathbf{v_1} = 2\hat{i} + 4\hat{j} - 5\hat{k} \) and \( \mathbf{v_2} = b\hat{i} + 2\hat{j} + 3\hat{k} \). The sum of these vectors is: \[ \mathbf{v_1} + \mathbf{v_2} = (2 + b)\hat{i} + (4 + 2)\hat{j} + (-5 + 3)\hat{k} = (2 + b)\hat{i} + 6\hat{j} - 2\hat{k} \] The unit vector in the direction of this sum is: \[ \hat{u} = \frac{\mathbf{v_1} + \mathbf{v_2}}{|\mathbf{v_1} + \mathbf{v_2}|} \] Now, the scalar product of \( \hat{i} + \hat{j} + \hat{k} \) and \( \hat{u} \) is given by: \[ (\hat{i} + \hat{j} + \hat{k}) \cdot \hat{u} = 1 \] After simplifying, we find that \( b = 1 \).