MHT CET · Maths · Three Dimensional Geometry
The distance of the point \((-1,-5,-10)\) from the point of intersection of the line \(\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}\) and the plane \(x-y+z=5\) is
- A 13 units.
- B 12 units.
- C 5 units.
- D 16 units.
Answer & Solution
Correct Answer
(A) 13 units.
Step-by-step Solution
Detailed explanation
Let \(\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}=\lambda\)
\(\therefore\) The co-ordinates of any point on the line are
\(
\mathrm{P} \equiv(3 \lambda+2,4 \lambda-1,12 \lambda+2)
\)
This point lies on the plane
\(x-y+z=5 \)
\( \therefore3 \lambda+2-(4 \lambda-1)+12 \lambda+2=5 \)
\( \Rightarrow 11 \lambda=0 \)
\( \Rightarrow \lambda=0 \)
\( \therefore \mathrm{P} \equiv(2,-1,2) \)
\( \text { Let } \mathrm{Q} \equiv(-1,-5,-10) \)
\( \therefore \mathrm{PQ}=\sqrt{(-1-2)^2+(-5+1)^2+(-10-2)^2} \)
\( =\sqrt{9+16+144} \)
\( =13 \text { units }\)
\(\therefore\) The co-ordinates of any point on the line are
\(
\mathrm{P} \equiv(3 \lambda+2,4 \lambda-1,12 \lambda+2)
\)
This point lies on the plane
\(x-y+z=5 \)
\( \therefore3 \lambda+2-(4 \lambda-1)+12 \lambda+2=5 \)
\( \Rightarrow 11 \lambda=0 \)
\( \Rightarrow \lambda=0 \)
\( \therefore \mathrm{P} \equiv(2,-1,2) \)
\( \text { Let } \mathrm{Q} \equiv(-1,-5,-10) \)
\( \therefore \mathrm{PQ}=\sqrt{(-1-2)^2+(-5+1)^2+(-10-2)^2} \)
\( =\sqrt{9+16+144} \)
\( =13 \text { units }\)
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