MHT CET · Maths · Three Dimensional Geometry
The equations of planes parallel to the plane \(x+2 y+2 z+8=0\), which are at a
distance of 2 units from the point \((1,1,2)\) are
- A \(x+2 y+2 z-13=0\) or \(x+2 y+2 z-1=0\)
- B \(x+2 y+2 z-6=0 \quad\) or \(\quad x+2 y+2 z-7=0\)
- C \(x+2 y+2 z+3=0 \quad\) or \(\quad x+2 y+2 z-5=0\)
- D \(x+2 y+2 z-5=0 \quad\) or \(\quad x+2 y+2 z-3=0\)
Answer & Solution
Correct Answer
(A) \(x+2 y+2 z-13=0\) or \(x+2 y+2 z-1=0\)
Step-by-step Solution
Detailed explanation
(B)
The equation of the plane parallel to the plane \(x+2 y+2 z+18=0\) is \(x+2 y+2 z+\lambda=0\). Now, the distance of this plane from the print \((1,1,2)\) is
\(\therefore\left|\frac{1(1)+2(1)+2(2)+\lambda}{\sqrt{1^{2}+2^{2}+2^{2}}}\right|=\left|\frac{7+\lambda}{3}\right|\)
\(\begin{array}{l}
\text { We have }\left|\frac{7+\lambda}{3}\right|=2 \Rightarrow \frac{7+\lambda}{3}=\pm 2 \\
\therefore \lambda=\pm 6-7=1 \Rightarrow \lambda=-1 \text { or } \lambda=-13
\end{array}\)
Hence, equations of plane are \(x+2 y+2 z-1=0\) or \(x+2 y+2 z-13=0\)
The equation of the plane parallel to the plane \(x+2 y+2 z+18=0\) is \(x+2 y+2 z+\lambda=0\). Now, the distance of this plane from the print \((1,1,2)\) is
\(\therefore\left|\frac{1(1)+2(1)+2(2)+\lambda}{\sqrt{1^{2}+2^{2}+2^{2}}}\right|=\left|\frac{7+\lambda}{3}\right|\)
\(\begin{array}{l}
\text { We have }\left|\frac{7+\lambda}{3}\right|=2 \Rightarrow \frac{7+\lambda}{3}=\pm 2 \\
\therefore \lambda=\pm 6-7=1 \Rightarrow \lambda=-1 \text { or } \lambda=-13
\end{array}\)
Hence, equations of plane are \(x+2 y+2 z-1=0\) or \(x+2 y+2 z-13=0\)
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