Question

If the points whose position vectors are $2 \hat{ i }+\hat{ j }+\hat{ k }, 6 \hat{ i }-\hat{ j }+2 \hat{ k }$ and $14 \hat{ i }-5 \hat{ j }+p \hat{ k }$ are collinear, then the value of $p$ is

• A. 2
• B. 4
• C. 6
• D. 8

Answer 1

Answer: (B)

Given vectors $\vec{ a }=2 \hat{ i }+\hat{ j }+\hat{ k }, \vec{ b }=6 \hat{ i }-\hat{ j }+2 \hat{ k }$
and $\vec{ c }=14 \hat{ i }-5 \hat{ j }+p \hat{ k }$ are collinear, therefore
$[\vec{ a } \vec{ b } \vec{ c }]=0$
$\Rightarrow \begin{vmatrix}2 & 1 & 1 \\ 6 & -1 & 2 \\ 14 & -5 & p\end{vmatrix}=0$
$\Rightarrow 2[-p+10]-1[6 p-28]+1[-30+14]=0$
$\Rightarrow -2 p+20-6 p+28-16=0$
$\Rightarrow -8 p+ 32=0$
$\Rightarrow p=4$