Question

The equation of the plane which passes through the line $a_{1}x+b_{1}y+c_{1}z+d_1=0,a_{2}x+b_{2}y+c_{2}z+d_2=0$ and which is parallel to the line $\frac{x-\alpha }{l}=\frac{y-\beta }{m}=\frac{z-\gamma }{n}$ is:

• A. $(a_{1}x+b_{1}y+c_{1}z+d_1)(a_{2}l+b_{2}m+c_{2}n)-(a_{1}l+b_{1}m+c_{1}n)(a_{2}x+b_{2}y+c_{2}z+d_2)=0$
• B. $(a_{1}x+b_{1}y+c_{1}z+d_1)(a_{2}l+b_{2}m+c_{2}n)+(a_{1}l+b_{1}m+c_{1}n)(a_{2}x+b_{2}y+c_{2}z+d_2)=0$
• C. $(a_{1}x+b_{1}y+c_{1}z-d_1)(a_{2}l+b_{2}m+c_{2}n)-(a_{1}l+b_{1}m+c_{1}n)(a_{2}x+b_{2}y+c_{2}z-d_2)=0$
• D. $(a_{1}x-b_{1}y-c_{1}z+d_1)(a_{2}l+b_{2}m+c_{2}n)-(a_{1}l+b_{1}m+c_{1}n)(a_{2}x-b_{2}y-c_{2}z+d_2)=0$

Answer 1

The correct option is A $(a_{1}x+b_{1}y+c_{1}z+d_1)(a_{2}l+b_{2}m+c_{2}n)-(a_{1}l+b_{1}m+c_{1}n)(a_{2}x+b_{2}y+c_{2}z+d_2)=0$

Given:$a_{1}x+b_{1}y+c_{1}z+d_1=0\text{and}{a}_{2}x+b_{2}y+c_{2}z+d_2=0\text{and}$
$\frac{x-\alpha }{l}=\frac{y-\beta }{m}=\frac{z-\gamma }{n}\cdots \left(i\right)$
Equation of palne through the inersection of the planes is given by $(a_{1}x+b_{1}y+c_{1}z+d_1)+\lambda (a_{2}x+b_{2}y+c_{2}z+d_2)=0$
$\Rightarrow (a_1+\lambda a_2)x+(b_1+\lambda b_2)y+(c_1+\lambda c_2)z+(d_1+\lambda d_2)=0\cdots \left(ii\right)$
$D.R^{\prime }s$ of the normal of the above equation of plane is $(a_1+\lambda a_2),(b_1+\lambda b_2),(c_1+\lambda c_2)$
Now, equation $\left(ii\right)$ is parallel to $\left(i\right)$
Hence normal to the plane in $\left(ii\right)$ must be perpendicular to line in equation $\left(i\right)$
$(a_1+\lambda a_2)l+(b_1+\lambda b_2)m+(c_1+\lambda c_2)n=0$
$\Rightarrow \lambda =-\frac{a_{1}l+b_{1}m+c_{1}n}{a_{2}l+b_{2}m+c_{2}n}$
Now, substituting it in equation $\left(ii\right)$, we have
$(a_{1}x+b_{1}y+c_{1}z+d_1)(a_{2}l+b_{2}m+c_{2}n)-(a_{1}l+b_{1}m+c_{1}n)(a_{2}x+b_{2}y+c_{2}z+d_2)=0$
This is the required equation of the plane.