Q71 / 160 · TS EAMCET 2024 (09 May Shift 1) · Mathematics
A vector of magnitude $\sqrt{2}$ units along the internal bisector of the angle between the vectors $2 \hat{i}-2 \hat{j}+\hat{k}$ and $\hat{i}+2 \hat{j}+2 \hat{k}$ is
- A$\hat{j}+\hat{k}$
- B$\hat{i}-\hat{j}$
- C$\hat{i}-\hat{k}$
- Correct answerD$\hat{i}+\hat{k}$
Solution
$\vec{a}=2 \hat{i}-2 \hat{j}+\hat{k}, \vec{b}=\hat{i}+2 \hat{j}+2 \hat{k}$
$|\vec{a}|=3=|\vec{b}|$
Internal angle bisector is $\frac{\vec{a}+\vec{b}}{2}=3 \hat{i}+3 \hat{k}$ Vector of magnitude $\sqrt{2}$ is $\hat{i}+\hat{k}$.
Internal angle bisector is $\frac{\vec{a}+\vec{b}}{2}=3 \hat{i}+3 \hat{k}$ Vector of magnitude $\sqrt{2}$ is $\hat{i}+\hat{k}$.
Hence, the correct answer is option: 'D'.