MHT CET · Maths · Three Dimensional Geometry
The perpendicular distance from the origin to the plane containing the two lines \(\frac{x+2}{3}=\frac{y-2}{5}=\frac{z+5}{7}\) and \(\frac{x-1}{1}=\frac{y-4}{4}=\frac{z+4}{7}\), is
- A \(\frac{11}{\sqrt{6}}\) units
- B \(11 \sqrt{6}\) units
- C 11 units
- D \(6 \sqrt{11}\) units
Answer & Solution
Correct Answer
(A) \(\frac{11}{\sqrt{6}}\) units
Step-by-step Solution
Detailed explanation
Equation of the plane is
\(\begin{aligned}
& \left|\begin{array}{ccc}
x-1 & y-4 & z+4 \\
3 & 5 & 7 \\
1 & 4 & 7
\end{array}\right| \\
& \Rightarrow(x-1)(7)-(y-4)(14)+(z+4)(7)=0 \\
& \Rightarrow 7 x-14 y+7 z+77=0 \\
& \Rightarrow x-2 y+z+11=0
\end{aligned}\)
\(\therefore \quad\) Perpendicular distance from origin to the plane is \(\left|\frac{0-2(0)+0+11}{\sqrt{1+4+1}}\right|=\frac{11}{\sqrt{6}}\) units
\(\begin{aligned}
& \left|\begin{array}{ccc}
x-1 & y-4 & z+4 \\
3 & 5 & 7 \\
1 & 4 & 7
\end{array}\right| \\
& \Rightarrow(x-1)(7)-(y-4)(14)+(z+4)(7)=0 \\
& \Rightarrow 7 x-14 y+7 z+77=0 \\
& \Rightarrow x-2 y+z+11=0
\end{aligned}\)
\(\therefore \quad\) Perpendicular distance from origin to the plane is \(\left|\frac{0-2(0)+0+11}{\sqrt{1+4+1}}\right|=\frac{11}{\sqrt{6}}\) units
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