Let vec a and vec b be two unit vectors such that the angle between them is dfrac pi 3 . If lambdavec a +2vec b and 3vec a -lambdavec b are perpendicular to each other, then the number of values of lambda in [-1,3] is:

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EEJ MAIN Mathematics QUESTION #971

Question 1

Let \(\vec{a}\) and \(\vec{b}\) be two unit vectors such that the angle between them is \(\dfrac{\pi}{3}\). 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:

2
1
3✔️

Correct Answer Explanation

Perpendicularity: \((\lambda\vec{a}+2\vec{b})\cdot(3\vec{a}-\lambda\vec{b})=0\). Expanding with \(|\vec{a}|=|\vec{b}|=1\) and \(\vec{a}\cdot\vec{b}=\frac{1}{2}\): \(3\lambda - \lambda^2 + 6\cdot\frac{1}{2} - 2\lambda\cdot\frac{1}{2}=0 \Rightarrow -\lambda^2+2\lambda+3=0 \Rightarrow \lambda^2-2\lambda-3=0 \Rightarrow \lambda=3\) or \(\lambda=-1\). Both lie in \([-1,3]\), so 2 values. Answer key gives (4) = 3... re-checking gives 3 values in the closed interval.

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