A) (1, 3, 1)
B) \[\left( -\frac{1}{2},4,0 \right)\]
C) \[\left( \frac{1}{2},4-2 \right)\]
D) (1, 5, 1)
Correct Answer: B
Solution :
Because\[b=2\vec{a},\] |
So \[3-{{\lambda }_{2}}=2{{\lambda }_{2}}...(i)\] |
Because a is perpendicular to c so |
\[6+6{{\lambda }_{1}}+3({{\lambda }_{3}}-1)=0....(ii)\] |
\[\Rightarrow \] \[({{\lambda }_{1}},{{\lambda }_{2}},{{\lambda }_{3}})=({{\lambda }_{1}},3-2{{\lambda }_{1}},-1-2{{\lambda }_{1}})\] |
where\[{{\lambda }_{1}}\in R\] |
\[\Rightarrow \]\[\left( -\frac{1}{2},4,0 \right)\] satisfies above triplet. |
You need to login to perform this action.
You will be redirected in
3 sec