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The angle between the pairs of lines

r⃗=(3i^+2j^−4k^)+λ(i^+2j^+2k^)\vec{r} = (3\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(\hat{i} + 2\hat{j} + 2\hat{k})

r⃗=(5i^−2j^)+μ(3i^+2j^+6k^)\vec{r} = (5\hat{i} - 2\hat{j}) + \mu(3\hat{i} + 2\hat{j} + 6\hat{k}) is :

Solution

✅ Correct Option: 3

Direction vectors: b1⃗=(1,2,2)\vec{b_1} = (1,2,2), b2⃗=(3,2,6)\vec{b_2} = (3,2,6). b1⃗⋅b2⃗=3+4+12=19\vec{b_1}\cdot\vec{b_2} = 3+4+12 = 19. ∣b1⃗∣=3|\vec{b_1}| = 3, ∣b2⃗∣=7|\vec{b_2}| = 7. cos⁡θ=1921\cos\theta = \frac{19}{21}. So θ=cos⁡−1(1921)\theta = \cos^{-1}\left(\frac{19}{21}\right).

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