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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 perpendicular is :

Solution

✅ Correct Option: 2

Perpendicular: (a⃗+λb⃗)⋅(a⃗−λb⃗)=0(\vec{a} + \lambda\vec{b}) \cdot (\vec{a} - \lambda\vec{b}) = 0, so ∣a⃗∣2−λ2∣b⃗∣2=0|\vec{a}|^2 - \lambda^2|\vec{b}|^2 = 0. Then 9−16λ2=09 - 16\lambda^2 = 0, giving λ2=916\lambda^2 = \frac{9}{16}, so λ=34\lambda = \frac{3}{4}.

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