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The angle between the vectors i^−j^\hat{i} - \hat{j} and j^−k^\hat{j} - \hat{k} is:

Solution

✅ Correct Option: 3

a⃗=i^−j^\vec{a} = \hat{i} - \hat{j}, b⃗=j^−k^\vec{b} = \hat{j} - \hat{k}. a⃗⋅b⃗=(1)(0)+(−1)(1)+(0)(−1)=−1\vec{a} \cdot \vec{b} = (1)(0) + (-1)(1) + (0)(-1) = -1. ∣a⃗∣=2|\vec{a}| = \sqrt{2}, ∣b⃗∣=2|\vec{b}| = \sqrt{2}. cos⁡θ=−12⋅2=−12\cos\theta = \frac{-1}{\sqrt{2}\cdot\sqrt{2}} = -\frac{1}{2}. So θ=2π/3\theta = 2\pi/3.

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