MHT CET · Maths · Three Dimensional Geometry
The distance of the point \((1,-5,9)\) from the plane \(x-y+z=5\) measured along the line \(x=y=z\) is ______ units.
- A \(3 \sqrt{10}\)
- B \(10 \sqrt{3}\)
- C \(\frac{10}{\sqrt{3}}\)
- D \(\frac{20}{3}\)
Answer & Solution
Correct Answer
(B) \(10 \sqrt{3}\)
Step-by-step Solution
Detailed explanation
Equation of the line can be written as \(\frac{x-1}{1}=\frac{y+5}{1}=\frac{z-9}{1}=\lambda\)
\(\therefore\) Co-ordinates of any point on the line are given as \(x=\lambda+1, y=\lambda-5, \mathrm{z}=\lambda+9\)
Substituting in the equation of the plane, we get
\((\lambda+1)-(\lambda-5)+(\lambda+9)=5 \)
\( \therefore \lambda+15=5 \)
\( \therefore \lambda=-10 \)
\( \therefore \text {Point on the plane is }(-9,-15,-1) \)
\( \therefore \text { Required distance } \)
\( =\text { Distance between }(1,-5,9) \text { and }(-9,-15,-1) \)
\( =\sqrt{(-9-1)^2+(-15+5)^2+(-1-9)^2} \)
\( =10 \sqrt{3}\)
\(\therefore\) Co-ordinates of any point on the line are given as \(x=\lambda+1, y=\lambda-5, \mathrm{z}=\lambda+9\)
Substituting in the equation of the plane, we get
\((\lambda+1)-(\lambda-5)+(\lambda+9)=5 \)
\( \therefore \lambda+15=5 \)
\( \therefore \lambda=-10 \)
\( \therefore \text {Point on the plane is }(-9,-15,-1) \)
\( \therefore \text { Required distance } \)
\( =\text { Distance between }(1,-5,9) \text { and }(-9,-15,-1) \)
\( =\sqrt{(-9-1)^2+(-15+5)^2+(-1-9)^2} \)
\( =10 \sqrt{3}\)
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