Vector Algebra
Perpendicularity condition on linear combinations of unit vectors
nta_pyq_2025_apr
Grade 12
Question:
Let $\hat{a}$ and $\hat{b}$ be two unit vectors such that the angle between them is $\dfrac{\pi}{3}$. If $\lambda\hat{a}+2\hat{b}$ and $3\hat{a}-\lambda\hat{b}$ are perpendicular to each other, then the number of values of $\lambda$ in $[-1,3]$ is:
Step-by-Step Solution
Key Concept: Set the dot product to zero using $\hat{a}\cdot\hat{b}=\cos(\pi/3)=\tfrac{1}{2}$, solve the resulting quadratic, and check whether the roots fall in $[-1,3]$.
$\hat{a}\cdot\hat{b}=\tfrac{1}{2}$. Perpendicularity: $(\lambda\hat{a}+2\hat{b})\cdot(3\hat{a}-\lambda\hat{b})=0$.
$3\lambda|\hat{a}|^2-\lambda^2\hat{a}\cdot\hat{b}+6\hat{a}\cdot\hat{b}-2\lambda|\hat{b}|^2=0$
$3\lambda-\dfrac{\lambda^2}{2}+3-2\lambda=0 \Rightarrow \lambda^2-2\lambda-6=0 \Rightarrow \lambda=1\pm\sqrt{7}$.
$1+\sqrt{7}\approx3.65>3$ and $1-\sqrt{7}\approx-1.65<-1$. Neither root lies in $[-1,3]$.
Number of values $=0$.
Correct Answer: 3