Vectors
Vectors
Allen Star Batch
Grade 12
Question:
$\vec{a}$ and $\vec{b}$ are two unit vectors inclined at an angle $\alpha(\alpha \in [0, \pi])$ to each other and $|\vec{a} + \vec{b}| < 1$ then $\alpha$ can lie in:
$\alpha \in \left[\frac{\pi}{3}, \frac{2\pi}{3}\right]$
$\alpha \in \left[\frac{\pi}{3}, \frac{\pi}{2}\right]$
$\alpha \in \left[\frac{2\pi}{3}, \frac{5\pi}{6}\right]$
$\alpha \in \left[\frac{2\pi}{3}, \frac{5\pi}{7}\right]$
Step-by-Step Solution
Key Concept: Use the formula $|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2|\vec{a}||\vec{b}|\cos\alpha$. Since both are unit vectors, $|\vec{a} + \vec{b}|^2 = 2 + 2\cos\alpha < 1$, which gives $\cos\alpha < -\frac{1}{2}$, implying $\alpha \in (\frac{2\pi}{3}, \pi]$.
Given $|\vec{a}+\vec{b}|0$ with $\vec{b}$ making obtuse angle, implying $\lambda<0$. Condition (C) requires solving the determinant $\begin{vmatrix}2 & -1 & 1 \\ 1 & 2 & 1+a \\ 3 & a & 5\end{vmatrix}=0$.
Correct Answer: 1,2