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Question: What is the equation of the plane passing through the line \(3x+y-5z=2=x-2y+3z\) and perpendicular t...

What is the equation of the plane passing through the line 3x+y−5z=2=x−2y+3z3x+y-5z=2=x-2y+3z and perpendicular to the plane x−y+z=3x-y+z=3.
(a) 2x+3y+z=22x+3y+z=2
(b) 3x+2y−z=23x+2y-z=2
(c) 7(x−z)=67\left( x-z \right)=6
(d) No plane exists

Explanation

Solution

For solving this question we will use the concept of the family of planes and form the equation of the required plane in terms of one variable and then write the vector perpendicular to the required plane. After that we will use the condition that two vectors will be perpendicular then their dot product will be zero.

Complete step-by-step solution
Given:
There is a line whose equation is 3x+y−5z=2=x−2y+3z3x+y-5z=2=x-2y+3z and we have to find the equation of a plane, which passes through this line and perpendicular to the plane x−y+z=3x-y+z=3.
Now, before we proceed we should know that ax+by+cz=dax+by+cz=d represents the general equation of a plane and a vector normal to this will be ai^+bj^+ck^a\widehat{i}+b\widehat{j}+c\widehat{k} .
Now, as the equation of the given line is 3x+y−5z=2=x−2y+3z3x+y-5z=2=x-2y+3z so, we can say that this line is the intersection of two planes whose equations are:
1. Equation of the first plane is P1:3x+y−5z−2=0{{P}_{1}}:3x+y-5z-2=0 .
2. Equation of the second plane is P2:x−2y+3z−2=0{{P}_{2}}:x-2y+3z-2=0 .
Now, the family of planes which passes through the intersection of the planes P1{{P}_{1}} and P2{{P}_{2}} will be P:P1+λP2=0P:{{P}_{1}}+\lambda {{P}_{2}}=0 where λ\lambda is a variable parameter. Then,
P:P1+λP2=0 ⇒P:3x+y−5z−2+λ(x−2y+3z−2)=0 ⇒P:(3+λ)x+(1−2λ)y+(3λ−5)z−2−2λ=0 \begin{aligned} & P:{{P}_{1}}+\lambda {{P}_{2}}=0 \\\ & \Rightarrow P:3x+y-5z-2+\lambda \left( x-2y+3z-2 \right)=0 \\\ & \Rightarrow P:\left( 3+\lambda \right)x+\left( 1-2\lambda \right)y+\left( 3\lambda -5 \right)z-2-2\lambda =0 \\\ \end{aligned}
Now, as there are infinite such planes which will pass through the intersection of the planes P1{{P}_{1}} and P2{{P}_{2}} . And the required plane will be one of them.
Now, let the equation of the required plane is P′:P1+λP2=0{P}':{{P}_{1}}+\lambda {{P}_{2}}=0 . Then,
P′:(3+λ)x+(1−2λ)y+(3λ−5)z−2−2λ=0{P}':\left( 3+\lambda \right)x+\left( 1-2\lambda \right)y+\left( 3\lambda -5 \right)z-2-2\lambda =0
Now, as we know that for a plane ax+by+cz=dax+by+cz=d vector ai^+bj^+ck^a\widehat{i}+b\widehat{j}+c\widehat{k} will be perpendicular to it. So, the vector perpendicular to the plane P′:(3+λ)x+(1−2λ)y+(3λ−5)z−2−2λ=0{P}':\left( 3+\lambda \right)x+\left( 1-2\lambda \right)y+\left( 3\lambda -5 \right)z-2-2\lambda =0 will be n′→=(3+λ)i^+(1−2λ)j^+(3λ−5)k^\overrightarrow{{{n}'}}=\left( 3+\lambda \right)\widehat{i}+\left( 1-2\lambda \right)\widehat{j}+\left( 3\lambda -5 \right)\widehat{k} . And the vector perpendicular to the plane x−y+z=3x-y+z=3 will be n1→=i^−j^+k^\overrightarrow{{{n}_{1}}}=\widehat{i}-\widehat{j}+\widehat{k} .
Now, as per the question, the plane P′:(3+λ)x+(1−2λ)y+(3λ−5)z−2−2λ=0{P}':\left( 3+\lambda \right)x+\left( 1-2\lambda \right)y+\left( 3\lambda -5 \right)z-2-2\lambda =0 and plane x−y+z=3x-y+z=3 are perpendicular so, the vectors perpendicular to these planes will be perpendicular to each other. In other words, vector n′→=(3+λ)i^+(1−2λ)j^+(3λ−5)k^\overrightarrow{{{n}'}}=\left( 3+\lambda \right)\widehat{i}+\left( 1-2\lambda \right)\widehat{j}+\left( 3\lambda -5 \right)\widehat{k} and vector n1→=i^−j^+k^\overrightarrow{{{n}_{1}}}=\widehat{i}-\widehat{j}+\widehat{k} will be perpendicular to each other so, their dot product will be zero. Then,
n′→⋅n1→=0 ⇒((3+λ)i^+(1−2λ)j^+(3λ−5)k^)⋅(i^−j^+k^)=0 ⇒3+λ−1+2λ+3λ−5=0 ⇒6λ−3=0 ⇒λ=0.5 \begin{aligned} & \overrightarrow{{{n}'}}\centerdot \overrightarrow{{{n}_{1}}}=0 \\\ & \Rightarrow \left( \left( 3+\lambda \right)\widehat{i}+\left( 1-2\lambda \right)\widehat{j}+\left( 3\lambda -5 \right)\widehat{k} \right)\centerdot \left( \widehat{i}-\widehat{j}+\widehat{k} \right)=0 \\\ & \Rightarrow 3+\lambda -1+2\lambda +3\lambda -5=0 \\\ & \Rightarrow 6\lambda -3=0 \\\ & \Rightarrow \lambda =0.5 \\\ \end{aligned}
Now, substitute λ=0.5\lambda =0.5 from the above equation in the equation of the plane P′:(3+λ)x+(1−2λ)y+(3λ−5)z−2−2λ=0{P}':\left( 3+\lambda \right)x+\left( 1-2\lambda \right)y+\left( 3\lambda -5 \right)z-2-2\lambda =0 . Then,
P′:(3+λ)x+(1−2λ)y+(3λ−5)z−2−2λ=0 ⇒P′:(3+0.5)x+(1−1)y+(1.5−5)z−2−1=0 ⇒P′:3.5x−3.5z−3=0 ⇒P′:3.5(x−z)=3 ⇒P′:7(x−z)=6 \begin{aligned} & {P}':\left( 3+\lambda \right)x+\left( 1-2\lambda \right)y+\left( 3\lambda -5 \right)z-2-2\lambda =0 \\\ & \Rightarrow {P}':\left( 3+0.5 \right)x+\left( 1-1 \right)y+\left( 1.5-5 \right)z-2-1=0 \\\ & \Rightarrow {P}':3.5x-3.5z-3=0 \\\ & \Rightarrow {P}':3.5\left( x-z \right)=3 \\\ & \Rightarrow {P}':7\left( x-z \right)=6 \\\ \end{aligned}
Thus, the equation of the required plane will be 7(x−z)=67\left( x-z \right)=6 .
Hence, (c) is the correct option.

Note: Here, the student should first understand what is asked in the problem and then proceed in the right direction. And we should make use of the concept of the family of planes to avoid calculation. Moreover, we should proceed in a stepwise manner and avoid calculation mistakes while solving.