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Let a⃗=(i^+4j^+2k^)\vec{a} = (\hat{i}+4\hat{j}+2\hat{k}), b⃗=(3i^−2j^+7k^)\vec{b} = (3\hat{i}-2\hat{j}+7\hat{k}), c⃗=(2i^−j^+4k^)\vec{c} = (2\hat{i}-\hat{j}+4\hat{k}). A vector d⃗\vec{d}, which is perpendicular to both a⃗\vec{a} and b⃗\vec{b}, such that c⃗.d⃗=18\vec{c}.\vec{d} = 18, is

Solution

✅ Correct Option: 2

a⃗×b⃗=32i^−j^−14k^\vec{a}\times\vec{b} = 32\hat{i} - \hat{j} - 14\hat{k}, so d⃗=λ(32i^−j^−14k^)\vec{d} = \lambda(32\hat{i} - \hat{j} - 14\hat{k}). Then c⃗⋅d⃗=λ(64+1−56)=9λ=18\vec{c}\cdot\vec{d} = \lambda(64 + 1 - 56) = 9\lambda = 18, so λ=2\lambda = 2 and d⃗=64i^−2j^−28k^\vec{d} = 64\hat{i} - 2\hat{j} - 28\hat{k}.

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