Browse Questions

# The vector $\hat i+\lambda\hat j+\hat k,\hat i+\hat j-\hat k$ and $3\hat i+6\hat j-5\hat k$ are coplanar if

$(A)\;\lambda=-2\quad (B)\;\lambda=0\quad(C)\;\lambda=1\quad (D)\;\lambda=-1$

Toolbox:
• If three vectors are coplanar, then $[a,b,c]=0$
Let $\overrightarrow a=\hat i+\lambda\hat j+\hat k,$ $\overrightarrow b=\hat i+\hat j-\hat k$ and $\overrightarrow c=3\hat i+6\hat j-5\hat k$
If three vectors are coplanar, then $[a,b,c]=0$
= $\begin{vmatrix} 1 & \lambda & 1 \\ 1 & 1 & -1 \\ 3 & 6 & -5 \end{vmatrix}$
On expanding we get,
$=1(-5+6)-\lambda(-5+3)+1(6-3)= 0$
On simplifying we get,
=>$1+2\lambda+3=0$
=>$2 \lambda=-4$
=>$\lambda=-2$
Hence $A$ is the correct option