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If a⃗=4i^+5j^−k^,b⃗=i^−4j^+5k^\vec a = 4\hat i+5\hat j-\hat k, \vec b=\hat i-4\hat j+5\hat k and c⃗=3i^+j^−k^\vec c=3\hat i+\hat j-\hat k. If a vector d⃗\vec d is perpendicular to both a⃗\vec a and b⃗\vec b and c⃗.d⃗=21\vec c.\vec d=21, then unit vector along d⃗\vec d is

Solution

✅ Correct Option: 2

Compute a⃗×b⃗=21i^−21j^−21k^=21(i^−j^−k^)\vec a \times \vec b=21\hat i-21\hat j-21\hat k=21(\hat i-\hat j-\hat k), so d⃗=λ(i^−j^−k^)\vec d=\lambda(\hat i-\hat j-\hat k). Then c⃗⋅d⃗=λ(3−1+1)=3λ=21\vec c \cdot \vec d=\lambda(3-1+1)=3\lambda=21, giving λ=7\lambda=7. The unit vector along d⃗\vec d is therefore 13(i^−j^−k^)\frac{1}{\sqrt{3}}(\hat i-\hat j-\hat k).

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