Question

The equation of the plane containing the $\frac{x-1}{3}=\frac{y-6}{4}=\frac{z-1}{2}$ and parallel to the line $\frac{x-2}{2}=\frac{y-1}{-3}=\frac{z+4}{3}$ is

• A. $18x-5y-17z+29=0$
• B. $18x-5y+17z+29=0$
• C. $26x-11y-17z-109=0$
• D. $26x-11y+17z-109=0$

Answer 1

The correct option is A $18x-5y-17z+29=0$

$\begin{array}{c}\frac{x-1}{3}-\frac{y-6}{4}=\frac{z-1}{2}\to \left(i\right)\\ \frac{x-2}{2}=\frac{y-1}{-3}=\frac{z+4}{3}\to \left(ii\right)\\ planepassesthroughline\left(1\right).henceplanewillpassthrough\left(1,6,1\right)\\ us,theequationofplaneis\\ a\left(x-1\right)+b\left(y-6\right)+c\left(z-1\right)=0\to \left(iii\right)\\ wherea,b,caredirectionratioofnormaltotheplane\\ now,\\ giventhatplaneisparalleltoline\left(2\right)\\ \therefore planeisparalleltobothline\left(1\right)and\left(2\right)\\ \to normalofplane=directionvectorofline\left(1\right)\times directionvectorofline\left(2\right)\\ \overrightarrow{n}=\left(3\hat{i}+4\hat{j}+2\hat{k}\right)\times \left(2\hat{i}-3\hat{j}+3\hat{k}\right)\\ =18\hat{i}-5\hat{j}-17\hat{k}\\ a=18,b=-5,c=-17\\ putthevaluesofa,b,cineq.\left(iii\right)\\ \Rightarrow 18\left(x-1\right)-5\left(y-6\right)-17\left(z-1\right)\\ =18x-5y-17z+29=0\end{array}$