MHT CET · Maths · Three Dimensional Geometry
The co-ordinates of the points on the line \(\frac{x+2}{1}=\frac{y-1}{2}=\frac{z+1}{-2}\) at a distance of 12 units from the point \(\mathrm{A}(-2,1,-1)\) are
- A \((2,9,-9),(-6,-7,7)\)
- B \((2,9,7),(6,5,-9)\)
- C \((6,9,-5),(-10,9,-5)\)
- D \((6,-7,3),(-10,9,3)\)
Answer & Solution
Correct Answer
(A) \((2,9,-9),(-6,-7,7)\)
Step-by-step Solution
Detailed explanation
\(\frac{\mathrm{x}+2}{1}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}+1}{-2}=\lambda \quad \ldots\) (say)
Hence coordinates of any point on the given line are \((\lambda-2,2 \lambda+1,-2 \lambda-1)\).
This point is at a distance of 12 units from \((-2,1,-1)\).
\(\therefore 12=\) \(\sqrt{(\lambda-2+2)^2+(2 \lambda+1-1)^2+(-2 \lambda-1+1)^2} \)
\( =\sqrt{\lambda^2+4 \lambda^2+4 \lambda^2}=3 \lambda \)
\( \therefore \lambda= \pm 4 \Rightarrow \text { Required point }=(2,9,-9) \text { or }\)\((-6,-7,7) \)
Hence coordinates of any point on the given line are \((\lambda-2,2 \lambda+1,-2 \lambda-1)\).
This point is at a distance of 12 units from \((-2,1,-1)\).
\(\therefore 12=\) \(\sqrt{(\lambda-2+2)^2+(2 \lambda+1-1)^2+(-2 \lambda-1+1)^2} \)
\( =\sqrt{\lambda^2+4 \lambda^2+4 \lambda^2}=3 \lambda \)
\( \therefore \lambda= \pm 4 \Rightarrow \text { Required point }=(2,9,-9) \text { or }\)\((-6,-7,7) \)
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