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Question 152 of 153

Q.If θ\theta is the angle between two vectors i^−2j^+k^\hat{i} - 2\hat{j} + \hat{k} and 3i^−2j^+k^3\hat{i} - 2\hat{j} + \hat{k}, find sin⁡θ\sin\theta.

Rajasthan RbseCBSE Class XII Board 2018Subjective· 2mImportance★★★★★
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sin⁡θ=10521\sin\theta=\dfrac{\sqrt{105}}{21}.

Concept. cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣\cos\theta=\dfrac{\vec a\cdot\vec b}{|\vec a||\vec b|}, and ∣a⃗×b⃗∣=∣a⃗∣∣b⃗∣sin⁡θ|\vec a\times\vec b|=|\vec a||\vec b|\sin\theta.

Why this method. The cross-product route gives sin⁡θ\sin\theta directly and avoids sign ambiguity.

Working. Let a⃗=i^−2j^+k^\vec a=\hat i-2\hat j+\hat k, b⃗=3i^−2j^+k^\vec b=3\hat i-2\hat j+\hat k.

∣a⃗∣=1+4+1=6,∣b⃗∣=9+4+1=14.|\vec a|=\sqrt{1+4+1}=\sqrt6,\qquad |\vec b|=\sqrt{9+4+1}=\sqrt{14}.

a⃗×b⃗=∣i^j^k^1−213−21∣=(0)i^−(−2)j^+(4)k^=2j^+4k^,\vec a\times\vec b=\begin{vmatrix}\hat i&\hat j&\hat k\\1&-2&1\\3&-2&1\end{vmatrix}=(0)\hat i-(-2)\hat j+(4)\hat k=2\hat j+4\hat k, …

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