A) 2
B) 4
C) 6
D) 8
Correct Answer: D
Solution :
\[\mathbf{a}+t\mathbf{b}=2\mathbf{i}+2\mathbf{j}+3\mathbf{k}+(-t\mathbf{i}+2t\mathbf{j}+t\mathbf{k})\] \[=(2-t)\mathbf{i}+(2+2t)\mathbf{j}+(3+t)\mathbf{k}\] Given that it is perpendicular to \[\mathbf{c}=3\mathbf{i}+\mathbf{j}\] Hence \[(2-t)3+(2+2t)1+(3+t)0=0\] \[\Rightarrow 6-3t+2+2t=0\Rightarrow t=8.\]You need to login to perform this action.
You will be redirected in
3 sec