Let \(\vec{a}=2\hat{\imath}+2\hat{\jmath}+\hat{k}\) and \(\vec{b}=6\hat{\imath}-3\hat{\jmath}-6\hat{k}\). A vector of magnitude \(3\sqrt{2}\) along the internal bisector of the angle between \(\vec{a}\) and \(\vec{b}\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(4\hat{\imath}+\hat{\jmath}-\hat{k}\)
The sum of two unit vectors is the diagonal of a rhombus, so it lies along the internal bisector of the angle between them.
\(|\vec{a}| = \sqrt{4+4+1} = 3, \quad |\vec{b}| = \sqrt{36+9+36} = 9\)
\( \displaystyle \hat{a}+\hat{b} = \frac{2\hat{\imath}+2\hat{\jmath}+\hat{k}}{3}+\frac{2\hat{\imath}-\hat{\jmath}-2\hat{k}}{3} = \frac{4\hat{\imath}+\hat{\jmath}-\hat{k}}{3} \)
\(|4\hat{\imath}+\hat{\jmath}-\hat{k}| = \sqrt{16+1+1} = 3\sqrt{2}\)
So \(4\hat{\imath}+\hat{\jmath}-\hat{k}\) already has magnitude \(3\sqrt{2}\) and points along the bisector.
Why the other options are incorrect
Option (1) \(8\hat{\imath}-\hat{\jmath}-5\hat{k}\) — Incorrect. This is \(\vec{a}+\vec{b}\); the vectors have different lengths, so their sum does not bisect the angle, and its length is \(\sqrt{90}\).
Option (2) \(3\hat{\jmath}+3\hat{k}\) — Incorrect. This is along \(\hat{a}-\hat{b} = \hat{\jmath}+\hat{k}\), the external bisector.
Option (4) \(4\hat{\imath}+\hat{\jmath}+\hat{k}\) — Incorrect. The z-component of \(\hat{a}+\hat{b}\) is \(\frac{1}{3}-\frac{2}{3} = -\frac{1}{3}\), so the sign of \(\hat{k}\) is wrong.