Question

Let \( \vec{a}, \vec{b}, \vec{c} \) be coinitial vectors and \( \vec{a} = 2\hat{i} - \hat{j} + 5\hat{k}, \vec{b} = 3\hat{i} + 7\hat{j} - \hat{k} \). If \( \cos(\theta) = 0 \), where \( \theta \) is the angle between \( \vec{a} \) and \( \vec{b} \), and \( \vec{c} \) is the vector along the bisector of the angle \( \angle ABC \), then the vector \( \mathbf{c} \) is:

• A. \( \lambda(5\hat{i} + 6\hat{j} + 4\hat{k}) \)
• B. \( \lambda(-\hat{i} - 8\hat{j} + 6\hat{k}) \)
• C. \( \lambda((2x + 3y)\hat{i} + (7y - x)\hat{j} + (5x - y)\hat{k}) \)
• D. \( \lambda((2x + 3y)\hat{i} + (x + 7y)\hat{j} + (5x + y)\hat{k}) \)

Hint:

For a vector along the angle bisector, the vector is simply a scalar multiple of the sum of the two vectors that form the angle.

Answer 1

The Correct Option is A

Step 1: Add the vectors \( \vec{a} \) and \( \vec{b} \).
\[ \vec{a} = 2\hat{i} - \hat{j} + 5\hat{k}, \quad \vec{b} = 3\hat{i} + 7\hat{j} - \hat{k} \] \[ \vec{a} + \vec{b} = (2 + 3)\hat{i} + (-1 + 7)\hat{j} + (5 - 1)\hat{k} = 5\hat{i} + 6\hat{j} + 4\hat{k} \]
Step 2: Express \( \vec{c} \) as a scalar multiple.
Since \( \mathbf{c} \) is along the bisector, we write: \[ \vec{c} = \lambda(5\hat{i} + 6\hat{j} + 4\hat{k}) \] Final Answer: \[ \boxed{\lambda(5\hat{i} + 6\hat{j} + 4\hat{k})} \]