MHT CET · Maths · Three Dimensional Geometry
The equation of the plane passing through the points \((2,3,1),(4,-5,3)\) and parallel
to \(\mathrm{y}\) -axis is
- A \(x+z=3\)
- B \(x+z=1\)
- C \(x-z=1\)
- D \(z-x+2=0\)
Answer & Solution
Correct Answer
(C) \(x-z=1\)
Step-by-step Solution
Detailed explanation
Equation of plane passing through the point \((2,3,1)\) is
\(a(x-2)+b(y-3)+c(z-1)=0\) \(\ldots(1)\)
Given point \((4,-5,3)\) lies on plane
\(2 a-8 b+2 c=0\) \(\ldots(2)\)
Since plane is parallel to \(\mathrm{Y}\)-axis, having d.r. \((0,1,0)\)
(a) \((0)+(\mathrm{b})(1)+(\mathrm{c})(0)=0 \Rightarrow \mathrm{b}=0\)
Putting in equation (2) we get
\(2 a+2 c=0 \Rightarrow a=-c\)
Putting values of \(\mathrm{a}, \mathrm{b}\) in equation (1)
\(\begin{aligned}
&-c(x-2)+c(z-1)=0 \\
\therefore &(x-2)-(z-1)=0 \\
\therefore & x-z=1
\end{aligned}\)
\(a(x-2)+b(y-3)+c(z-1)=0\) \(\ldots(1)\)
Given point \((4,-5,3)\) lies on plane
\(2 a-8 b+2 c=0\) \(\ldots(2)\)
Since plane is parallel to \(\mathrm{Y}\)-axis, having d.r. \((0,1,0)\)
(a) \((0)+(\mathrm{b})(1)+(\mathrm{c})(0)=0 \Rightarrow \mathrm{b}=0\)
Putting in equation (2) we get
\(2 a+2 c=0 \Rightarrow a=-c\)
Putting values of \(\mathrm{a}, \mathrm{b}\) in equation (1)
\(\begin{aligned}
&-c(x-2)+c(z-1)=0 \\
\therefore &(x-2)-(z-1)=0 \\
\therefore & x-z=1
\end{aligned}\)
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