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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