Let O be the origin and the position vector of A and B be 2 i+2 j+ k$2\hat{i}+2\hat{j}+\hat{k}$ and 2 i+4 j+4 k$2\hat{i}+4\hat{j}+4\hat{k}$ respectively. If the internal bisector of ∠ AOB$\angle AOB$ meets the line AB at C, then the length of OC is
Vector angle bisector questions invariably test the geometric property that the bisector divides the opposite side in the ratio of the side lengths. Combine this directly with the 3D coordinate section formula.
Let a$\vec{a}$ and b$\vec{b}$ be two vectors such that | a| = 1, | b| = 4$|\vec{a}| = 1, |\vec{b}| = 4$ and a · b = 2$\vec{a} \cdot \vec{b} = 2$. If c = (2 a × b) - 3 b$\vec{c} = (2\vec{a} \times \vec{b}) - 3\vec{b}$ and the angle between b$\vec{b}$ and c$\vec{c}$ is α$\alpha$, then 192 ²α$192\sin^2\alpha$ is equal to
Numerical Answer.Answer: 48 to 48
Solution
Core Logic
b · c = b · ((2 a × b) - 3 b)$$\vec{b} \cdot \vec{c} = \vec{b} \cdot ((2\vec{a} \times \vec{b}) - 3\vec{b})$$|b||c| α = 2( b · ( a × b)) - 3|b|²$$|b||c|\cos\alpha = 2(\vec{b} \cdot (\vec{a} \times \vec{b})) - 3|b|^2$$
Since b · ( a × b) = 0$\vec{b} \cdot (\vec{a} \times \vec{b}) = 0$, we have |b||c| α = -3|b|²$|b||c|\cos\alpha = -3|b|^2$.
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