A) ? 2
B) ? 1
C) 0
D) 1
Correct Answer: D
Solution :
Parallel vector \[=(2+b)\mathbf{i}+6\mathbf{j}-2\mathbf{k}\] Unit vector \[=\frac{(2+b)\mathbf{i}+6\mathbf{j}-2\mathbf{k}}{\sqrt{{{b}^{2}}+4b+44}}\] According to the condition, \[1=\frac{(2+b)+6-2}{\sqrt{{{b}^{2}}+4b+44}}\] \[\Rightarrow {{b}^{2}}+4b+44={{b}^{2}}+12b+36\]\[\Rightarrow 8b=8\Rightarrow b=1.\]You need to login to perform this action.
You will be redirected in
3 sec