Question

If $A=2i+2j+4k,B=-i+2j+k$ and $C=3i+j,$ then $A+tB$ is perpendicular to $C$, if $t$ is equal to

• A. $8$
• B. $4$
• C. $6$
• D. $2$

Answer 1

Answer: (A)

$8$

$\overrightarrow{A}+t\overrightarrow{B}=(2-t)i+(2+2t)j+(4+t)k$
Since it is perpendicular to $\overrightarrow{C}$, hence their dot products will be 0.

Therefore

$(\overrightarrow{A}+t\overrightarrow{B}).\overrightarrow{C}=0$
$\left[\right(2-t)i+(2+2t)j+(4+t\left)k\right].[3i+j+0k]=0$

$\Rightarrow 3(2-t)+(2+2t)=0$
$\Rightarrow 6-3t+2+2t=0$
$\Rightarrow 8-t=0$
$\Rightarrow t=8$