MHT CET · Maths · Three Dimensional Geometry
The co-ordinates of the foot of the perpendicular from the point \((0,2,3)\) on the line
\(\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}\) are
- A \((-2,-3,1)\)
- B \((2,3,-1)\)
- C \((2,3,1)\)
- D (-2,-3,-1)
Answer & Solution
Correct Answer
(B) \((2,3,-1)\)
Step-by-step Solution
Detailed explanation
Let \(\alpha\) be the foot of the Ler drawn from the point \(\mathrm{P}(0,2,3)\) to the given line.
Co-ordinates of any point on given line are \(\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}=\lambda\)...say
Let \(Q \equiv(5 \lambda-3,2 \lambda+1,3 \lambda-4)\) and \(P \equiv(0,2,3)\)
\(\therefore\) d.r. of PQ are \(5 \lambda-3,2 \lambda-1,3 \lambda-7\)
d.r. of given line are \(5,2,3\)
\(\therefore 5(5 \lambda-3)+2(2 \lambda-1)+3(3 \lambda-7)=0\)
\(\Rightarrow 25 \lambda-15+4 \lambda-2+9 \lambda-21=0 \Rightarrow 38 \lambda-38=\) \(0 \Rightarrow \lambda=1\)
\(\therefore \mathrm{Q} \equiv(2,1,-4)\)

Co-ordinates of any point on given line are \(\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}=\lambda\)...say
Let \(Q \equiv(5 \lambda-3,2 \lambda+1,3 \lambda-4)\) and \(P \equiv(0,2,3)\)
\(\therefore\) d.r. of PQ are \(5 \lambda-3,2 \lambda-1,3 \lambda-7\)
d.r. of given line are \(5,2,3\)
\(\therefore 5(5 \lambda-3)+2(2 \lambda-1)+3(3 \lambda-7)=0\)
\(\Rightarrow 25 \lambda-15+4 \lambda-2+9 \lambda-21=0 \Rightarrow 38 \lambda-38=\) \(0 \Rightarrow \lambda=1\)
\(\therefore \mathrm{Q} \equiv(2,1,-4)\)

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