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 \[\vec{b}\]. |
You need to login to perform this action.
You will be redirected in
3 sec