Question

The Cartesian equation of the plane passing through A and perpendicular to $\overline{AB}$ where $3i+j+2k,i-2j-4k$ are the position vectors of A,B respectively

• A. $2x+3y-2z-5=0$
• B. $2x-3y+6z-21=0$
• C. $2x+3y+6z+21=0$
• D. $2x+3y+6z-21=0$

Answer 1

The correct option is D $2x+3y+6z-21=0$

$\bullet \overline{A}=3\hat{i}+\hat{j}+2\hat{k},\overline{B}=\hat{i}-2\hat{j}-4\hat{k}$

$\bullet \overline{AB}=2\hat{i}+3\hat{j}+6\hat{k}$

Thus normal vector of the plane is $\overline{A}B=2\hat{i}+3\hat{j}+6\hat{k}and\overline{A}$ lies on plane

Thus eqn of plane

$\overline{N}.(\overline{X}-\overline{A})=0where\overline{N}=\overline{A}B,\overline{X}=x\overline{i}+y\overline{i}+z\overline{k},$

$2x+3y+6z-(2\hat{i}+3\hat{j}+6\hat{k}).(3\hat{i}+\hat{j}+2\hat{k})=0$

$2x+3y+6z-21=0$