Question

a plane containing the point ( 3,2,0) and the line $\frac{x-1}{1}=\frac{y-2}{5}=\frac{z-3}{4}$ also containes the point

• A. (0,-3, 1)
• B. (0, 7 ,10)
• C. (0, 7, -10)
• D. (0 , 3, 1)

Answer 1

The correct option is B (0, 7 ,10)

Plane contains point $A(3,2,0)$ and also $B(1,2,3)$ {line point}

$\Rightarrow$ it is parallel to $\overrightarrow{AB}=-2\hat{i}+3\hat{k}$ also it is parallel to $\overrightarrow{l}=\hat{i}+5\hat{j}+4\hat{k}$

$\Rightarrow$ its normal vector would be parallel to $\overrightarrow{AB}\times \overrightarrow{l}$

$=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -2& 0& 3\\ 1& 5& 4\end{array}\right|=-15\hat{i}+\left(11\right)\hat{j}-10\hat{k}$

$\Rightarrow$ Equation of plane = $(\overrightarrow{r}-\overrightarrow{a}).\overrightarrow{n}=0$

$\Rightarrow \overrightarrow{r}.\overrightarrow{n}=\overrightarrow{a}.\overrightarrow{n}$

$\Rightarrow \overrightarrow{r}-(-15\hat{i}+11\hat{j}-10\hat{k})=-45+22$

$=-23$

$\Rightarrow -15x+11y-10z=-23$

$\left(A\right)-33-10\pm -23$

$\left(B\right)77-100=-23$ (B)