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If ∣a⃗∣=3|\vec{a}| = 3 and ∣b⃗∣=4|\vec{b}| = 4, then a value of λ\lambda for which a⃗+λb⃗\vec{a} + \lambda\vec{b} and a⃗−λb⃗\vec{a} - \lambda\vec{b} are perpendiculars is.

Solution

✅ Correct Option: 2

Perpendicularity gives (a⃗+λb⃗)⋅(a⃗−λb⃗)=∣a⃗∣2−λ2∣b⃗∣2=0(\vec{a}+\lambda\vec{b})\cdot(\vec{a}-\lambda\vec{b}) = |\vec{a}|^2 - \lambda^2|\vec{b}|^2 = 0. So 9=16λ29 = 16\lambda^2 and λ=34\lambda = \frac{3}{4}.

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