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Q.Let a⃗=4i^+5j^−k^\vec{a} = 4\hat{i} + 5\hat{j} - \hat{k}, b⃗=i^−4j^+5k^\vec{b} = \hat{i} - 4\hat{j} + 5\hat{k} and c⃗=3i^+j^−k^\vec{c} = 3\hat{i} + \hat{j} - \hat{k}. Find a vector d⃗\vec{d} which is perpendicular to both c⃗\vec{c} and b⃗\vec{b} and d⃗⋅a⃗=21\vec{d} \cdot \vec{a} = 21.

Punjab PsebCBSE Class XII Board 2018Subjective· 4mImportance★★★★★
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d⃗=13(−i^+16j^+13k^)\vec d=\dfrac13\left(-\hat i+16\hat j+13\hat k\right).

Concept. The cross product b⃗×c⃗\vec b\times\vec c is perpendicular to both b⃗\vec b and c⃗\vec c.

Why this method. Every vector perpendicular to both is a scalar multiple of b⃗×c⃗\vec b\times\vec c; the scalar is pinned by the dot-product condition.

Working. b⃗=(1,−4,5), c⃗=(3,1,−1)\vec b=(1,-4,5),\ \vec c=(3,1,-1).

b⃗×c⃗=∣i^j^k^1−4531−1∣=(−1)i^+(16)j^+(13)k^.\vec b\times\vec c=\begin{vmatrix}\hat i&\hat j&\hat k\\1&-4&5\\3&1&-1\end{vmatrix}=(-1)\hat i+(16)\hat j+(13)\hat k.

Let d⃗=t(−i^+16j^+13k^)\vec d=t(-\hat i+16\hat j+13\hat k). With a⃗=(4,5,−1)\vec a=(4,5,-1), …

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