MHT CET · Maths · Three Dimensional Geometry
The equation of the plane containing the line \(\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}\) and the point \((0,7,-7)\) is
- A \(x+y+z=1\)
- B \(x+y+z=2\)
- C \(x+y+z=0\)
- D None of these
Answer & Solution
Correct Answer
(C) \(x+y+z=0\)
Step-by-step Solution
Detailed explanation
The equation of plane containing the line \(\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}\) is
\(
a(x+1)+b(y-3)+c(z+2)=0 ...(i)
\)
where \(\quad-3 a+2 b+c=0\) ...(ii)
This passes through \((0,7,-7)\).
\(\therefore\) \(a+4 b-5 c=0\) ...(iii)
From Eqs. (ii) and (iii),
or
\(
\begin{aligned}
\frac{a}{-14} &=\frac{b}{-14}=\frac{c}{-14} \\
\frac{a}{1} &=\frac{b}{1}=\frac{c}{1}
\end{aligned}
\)
Thus, the required plane is \(x+y+z=0\)
\(
a(x+1)+b(y-3)+c(z+2)=0 ...(i)
\)
where \(\quad-3 a+2 b+c=0\) ...(ii)
This passes through \((0,7,-7)\).
\(\therefore\) \(a+4 b-5 c=0\) ...(iii)
From Eqs. (ii) and (iii),
or
\(
\begin{aligned}
\frac{a}{-14} &=\frac{b}{-14}=\frac{c}{-14} \\
\frac{a}{1} &=\frac{b}{1}=\frac{c}{1}
\end{aligned}
\)
Thus, the required plane is \(x+y+z=0\)
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