Vector Algebra
Vectors
star_batch_jee_advanced_2025
Grade None
Question:
A vector $\vec{c}$, directed along the internal bisector of the angle between the vectors $\vec{a}=7\vec{i}-4\vec{j}-4\vec{k}$ and $\vec{b}=-2\vec{i}-\vec{j}+2\vec{k}$ with $|\vec{c}|=5\sqrt{6}$, is :
5/3(\vec{i}-7\vec{j}+2\vec{k})
5/3(5\vec{i}+5\vec{j}+2\vec{k})
5/3(\vec{i}+7\vec{j}+2\vec{k})
5/3(-5\vec{i}+5\vec{j}+2\vec{k})
Step-by-Step Solution
Key Concept: The bisector is found by adding unit vectors along each direction and scaling by the required magnitude.
The vector along the bisector of the given angle is $\vec{c} = t\left(\frac{\vec{i} - 4\vec{j} - 4\vec{k}}{\sqrt{81}} + \frac{-2\vec{i} - \vec{j} + 2\vec{k}}{\sqrt{9}}\right) = t\left(\frac{\vec{i} - 7\vec{j} + 2\vec{k}}{9}\right)$. Since $|\vec{c}| = 5\sqrt{6} \cdot 5\sqrt{6} = \frac{\sqrt{54}}{9} \Rightarrow t = 15$.
Correct Answer: 1