Let $\vec{a}$ and $\vec{b}$ be two unit vectors such that the angle between them is $\frac{\pi}{3}$. If $\lambda\vec{a}+2\vec{b}$ and $3\vec{a}-\lambda\vec{b}$ are perpendicular to each other, then the number of values of $\lambda$ in $[-1, 3]$ is:
- (1) 3
- (2) 2
- (3) 1
- (4) 0
Solution:
Given $|\vec{a}| = 1, |\vec{b}| = 1$, and angle $\theta = \frac{\pi}{3}$.
$\vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos(\frac{\pi}{3}) = 1 \cdot 1 \cdot \frac{1}{2} = \frac{1}{2}$.
Since the vectors are perpendicular, their dot product is 0:
$$ (\lambda\vec{a} + 2\vec{b}) \cdot (3\vec{a} – \lambda\vec{b}) = 0 $$
$$ 3\lambda(\vec{a}\cdot\vec{a}) – \lambda^2(\vec{a}\cdot\vec{b}) + 6(\vec{b}\cdot\vec{a}) – 2\lambda(\vec{b}\cdot\vec{b}) = 0 $$
Substitute $|\vec{a}|^2 = 1, |\vec{b}|^2 = 1$ and $\vec{a}\cdot\vec{b} = \frac{1}{2}$:
$$ 3\lambda(1) – \lambda^2\left(\frac{1}{2}\right) + 6\left(\frac{1}{2}\right) – 2\lambda(1) = 0 $$
$$ 3\lambda – \frac{\lambda^2}{2} + 3 – 2\lambda = 0 $$
$$ \lambda – \frac{\lambda^2}{2} + 3 = 0 $$
Multiply by -2 to simplify:
$$ \lambda^2 – 2\lambda – 6 = 0 $$
$$ \lambda = \frac{2 \pm \sqrt{4 – 4(1)(-6)}}{2} = \frac{2 \pm \sqrt{28}}{2} = 1 \pm \sqrt{7} $$
Approximate values ($\sqrt{7} \approx 2.64$):
$\lambda_1 \approx 3.64$ (Not in $[-1, 3]$)
$\lambda_2 \approx -1.64$ (Not in $[-1, 3]$)
Thus, the number of values in $[-1, 3]$ is 0.
Ans. (4)