Question

A vector of magnitude \(\sqrt{2}\) units along the internal bisector of the angle between the vectors \(2\mathbf{i} - 2\mathbf{j} + \mathbf{k}\) and \(\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}\) is:

• A. \(\mathbf{j} + \mathbf{k}\)
• B. \(\mathbf{i} - \mathbf{j}\)
• C. \(\mathbf{i} - \mathbf{k}\)
• D. \(\mathbf{i} + \mathbf{k}\)

Hint:

When calculating vectors along the angle bisector, ensure normalization to unit length before applying the magnitude scaling.

Answer 1

The Correct Option is D

Let \(\mathbf{a} = 2\mathbf{i} - 2\mathbf{j} + \mathbf{k}\) and \(\mathbf{b} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k}\). We need to find the unit vectors along \(\mathbf{a}\) and \(\mathbf{b}\): \[\hat{\mathbf{a}} = \frac{\mathbf{a}}{|\mathbf{a}|} = \frac{2\mathbf{i} - 2\mathbf{j} + \mathbf{k}}{\sqrt{2^2 + (-2)^2 + 1^2}} = \frac{2\mathbf{i} - 2\mathbf{j} + \mathbf{k}}{\sqrt{9}} = \frac{2\mathbf{i} - 2\mathbf{j} + \mathbf{k}}{3}\] \[\hat{\mathbf{b}} = \frac{\mathbf{b}}{|\mathbf{b}|} = \frac{\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}}{\sqrt{1^2 + 2^2 + 2^2}} = \frac{\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}}{\sqrt{9}} = \frac{\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}}{3}\] The vector along the internal bisector of the angle between \(\mathbf{a}\) and \(\mathbf{b}\) is given by \(\hat{\mathbf{a}} + \hat{\mathbf{b}}\). \[\hat{\mathbf{a}} + \hat{\mathbf{b}} = \frac{2\mathbf{i} - 2\mathbf{j} + \mathbf{k}}{3} + \frac{\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}}{3} = \frac{3\mathbf{i} + 3\mathbf{k}}{3} = \mathbf{i} + \mathbf{k}\] We need a vector of magnitude \(\sqrt{2}\) along \(\mathbf{i} + \mathbf{k}\). The magnitude of \(\mathbf{i} + \mathbf{k}\) is \(\sqrt{1^2 + 1^2} = \sqrt{2}\). So, the vector of magnitude \(\sqrt{2}\) along the internal bisector is \(\mathbf{i} + \mathbf{k}\). Final Answer: The final answer is $\boxed{(4)}$