A) \[\vec{a}\cdot \hat{i}+3=0\]
B) \[\vec{a}\cdot \hat{i}+1=0\]
C) \[\vec{a}\cdot \hat{k}+2=0\]
D) \[\vec{a}\cdot \hat{k}+4=0\]
Correct Answer: C
Solution :
| [c] |
| \[\vec{a}={{\lambda }_{1}}(\hat{b}+\hat{c})\] |
| \[={{\lambda }_{1}}\left( \frac{\hat{i}+\hat{j}}{\sqrt{2}}+\frac{\hat{i}-\hat{j}+4\hat{k}}{3\sqrt{2}} \right)\] |
| \[=\frac{{{\lambda }_{1}}}{3\sqrt{2}}\left( 4\hat{i}+2\hat{j}+4\hat{k} \right)\] |
| As \[\vec{a}=\alpha \hat{i}+2\hat{j}+\beta \hat{k}\] |
| \[\therefore {{\lambda }_{1}}=3\sqrt{2},\,\,\,\,\alpha =4,\beta =4\] |
| \[\vec{a}=4\hat{i}+2\hat{j}+4\hat{k}\] no option is satisfied |
| Also \[\vec{a}={{\lambda }_{2}}(\hat{b}-\hat{c})\] |
| \[=\frac{{{\lambda }_{2}}}{3\sqrt{2}}\left( (3\hat{i}+3\hat{j})-(\hat{i}-\hat{j}+4\hat{k}) \right)=\frac{{{\lambda }_{2}}}{3\sqrt{2}}(2\hat{i}+4\hat{j}-4\hat{k})\] |
| \[=\frac{2{{\lambda }_{2}}}{3\sqrt{2}}\left( \hat{i}+2\hat{j}-2\hat{k} \right)\] |
| \[\Rightarrow \alpha =1,\beta =-2\] and \[\frac{2{{\lambda }_{2}}}{3\sqrt{2}}=1\] |
| \[\therefore \vec{a}\cdot \hat{k}+2=0\] |
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