Question

The plane through the intersection of the planes $x+y+z-1=0$ and $2x+3y-z+4=0$ and parallel to the y-axis passes through the point.
(a) $\left( -3,0,1 \right)$
(b) $\left( 3,3,-1 \right)$
(c) $\left( 3,2,1 \right)$
(d) $\left( -3,1,1 \right)$

Answer 1

Hint: Consider a general plane ${{P}_{1}}+\lambda {{P}_{2}}=0$ passing through the point of intersection of the plane ${{P}_{1}}=x+y+z-1$ and ${{P}_{2}}=2x+3y-z+4$ . Now we know that for a general plane ax+by+cz+d=0, $a\widehat{i}+b\widehat{j}+c\widehat{k}$ represents the vector normal to the given plane and it is given that the plane is parallel to y-axis, so the dot product of the normal vector of the plane in terms of $\lambda$ and vector $\widehat{j}$ is equal to zero. So, using this find the plane and put the options to get the answer.

Complete step-by-step answer:
Let us start by considering the unknown plane to be ${{P}_{1}}+\lambda {{P}_{2}}=0$ , where ${{P}_{1}}=x+y+z-1$ and ${{P}_{2}}=2x+3y-z+4$ . We are allowed to do so as it is given that the plane pass through the intersection of the planes ${{P}_{1}}=x+y+z-1$ and ${{P}_{2}}=2x+3y-z+4$ .
So, the plane comes out to be:
${{P}_{1}}+\lambda {{P}_{2}}=0$
$\Rightarrow x+y+z-1+\lambda \left( 2x+3y-z+4 \right)=0$
$\Rightarrow \left( 2\lambda +1 \right)x+\left( 3\lambda +1 \right)y+\left( 1-\lambda \right)z-1+4\lambda =0$
Now we know that for a general plane ax+by+cz+d=0, $a\widehat{i}+b\widehat{j}+c\widehat{k}$ represents the vector normal to the given plane.
Therefore $\left( 2\lambda +1 \right)\widehat{i}+\left( 3\lambda +1 \right)\widehat{j}+\left( 1-\lambda \right)\widehat{k}$ is normal to $\left( 2\lambda +1 \right)x+\left( 3\lambda +1 \right)y+\left( 1-\lambda \right)z-1+4\lambda =0$ . Also, it is given that the y-axis, i.e., $\widehat{j}$ is parallel to the plane, hence, normal to $\left( 2\lambda +1 \right)\widehat{i}+\left( 3\lambda +1 \right)\widehat{j}+\left( 1-\lambda \right)\widehat{k}$ vector, and we know that the dot product of two normal vectors is equal to zero.
$\begin{aligned} & \left( \left( 2\lambda +1 \right)\widehat{i}+\left( 3\lambda +1 \right)\widehat{j}+\left( 1-\lambda \right)\widehat{k} \right).\widehat{j}=0 \\\ & \Rightarrow 3\lambda +1=0 \\\ & \Rightarrow \lambda =\dfrac{-1}{3} \\\ \end{aligned}$
Now, if we put this value in the equation of plane, we get the equation of plane to be:
$\left( 2\lambda +1 \right)x+\left( 3\lambda +1 \right)y+\left( 1-\lambda \right)z-1+4\lambda =0$

& \Rightarrow \left( 2\times \left( \dfrac{-1}{3} \right)+1 \right)x+\left( 3\times \left( \dfrac{-1}{3} \right)+1 \right)y+\left( 1-\left( \dfrac{-1}{3} \right) \right)z-1+4\times \left( \dfrac{-1}{3} \right)=0 \\\ & \Rightarrow \dfrac{1}{3}x+0y+\dfrac{4}{3}z-\dfrac{7}{3}=0 \\\ & \Rightarrow x+4z-7=0 \\\ \end{aligned}$$ Now let us put the options one-by-one in the equation of the plane to get the answer. Let us start with $\left( -3,0,1 \right)$ . $$\begin{aligned} & -3+4\times 1-7\ne 0 \\\ & \Rightarrow -6\ne 0 \\\ \end{aligned}$$ So, the equation is not satisfied. Now let us check for $\left( 3,3,-1 \right)$ . $$\begin{aligned} & 3+4\times \left( -1 \right)-7\ne 0 \\\ & \Rightarrow -8\ne 0 \\\ \end{aligned}$$ This is also not satisfying the equation. Now let us check for $\left( 3,2,1 \right)$ . $$\begin{aligned} & 3+4\times 1-7=0 \\\ & \Rightarrow 0=0 \\\ \end{aligned}$$ So, the equation is satisfied for $\left( 3,2,1 \right)$ . Hence, the answer is option (c). Note: Be careful and don’t confuse the statement that the plane passes through the point of intersection with the statement that the plane bisects the two given plane and directly use the formula of bisector plane of two given planes, i.e. $\dfrac{{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}}=\pm \dfrac{{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}}}{\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}$ . Also, try to keep the equations as simple as possible to reduce the chances of making errors.