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Let a⃗=i^−j^\vec{a}=\hat{i}-\hat{j}, b⃗=3j^−k^\vec{b}=3\hat{j}-\hat{k} and c⃗=7i^−k^\vec{c}=7\hat{i}-\hat{k}. Then a vector d⃗\vec{d} which is perpendicular to both a⃗\vec{a} and b⃗\vec{b} and c⃗.d⃗=1\vec{c}.\vec{d}=1 is

Solution

✅ Correct Option: 2

a⃗×b⃗=i^+j^+3k^\vec a\times\vec b=\hat i+\hat j+3\hat k, so d⃗=λ(i^+j^+3k^)\vec d=\lambda(\hat i+\hat j+3\hat k). Then c⃗⋅d⃗=λ(7+0−3)=4λ=1\vec c\cdot\vec d=\lambda(7+0-3)=4\lambda=1, giving λ=14\lambda=\dfrac14.

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