A) \[-2\]
B) \[-1\]
C) \[0\]
D) \[1\]
Correct Answer: D
Solution :
Parallel vector\[=(2+b)\widehat{\mathbf{i}}+6\widehat{\mathbf{j}}-2\widehat{\mathbf{k}}\] Unit vector\[=\frac{(2+b)\widehat{\mathbf{i}}+6\widehat{\mathbf{j}}-2\widehat{\mathbf{k}}}{\sqrt{{{b}^{2}}+4b+44}}\] According to the question, \[1=\frac{(2+b)+6-2}{\sqrt{{{b}^{2}}+4b+44}}\] \[\Rightarrow \] \[{{b}^{2}}+4b+44={{b}^{2}}+12b+36\] \[\Rightarrow \] \[8b=8\Rightarrow b=1\]
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