A) -1
B) \[\frac{1}{2}\]
C) \[-\frac{1}{2}\]
D) 1
Correct Answer: C
Solution :
Key Idea: Two vectors must be perpendicular if their dot product is zero. Let \[\vec{a}=2\hat{i}+3\hat{j}+8\hat{k}\] \[\vec{b}=4\hat{j}-4\hat{i}+\alpha \hat{k}\] \[=-4\hat{i}+4\hat{j}+\alpha \hat{k}\] According to the above hypothesis: \[\vec{a}\bot \,\vec{b}\] \[\Rightarrow \vec{a}\,.\vec{b}=0\] \[\Rightarrow \,\,(2\hat{i}+3\hat{j}+8\hat{k})\,(-4\hat{k}+4\hat{j}+\alpha \hat{k})=0\] \[\Rightarrow \,\,\,-8+12+8\alpha =0\] \[\Rightarrow 8\alpha =-4\] \[\therefore \alpha =-\frac{4}{8}=-\frac{1}{2}\] Note: \[\vec{a}.\vec{b}=ab\,\cos \theta \]. Here, a and b are always positive as they are the magnitudes of \[\vec{a}\] and \[\vec{b}\].
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