MHT CET · Maths · Three Dimensional Geometry
The coordinates of the foot of the perpendicular drawn from the origin to the plane \(2 x+y-2 z=18\) are
- A \((4,2,-4)\)
- B \((1,2,-3)\)
- C \((4,2,4)\)
- D \((4,-2,-4)\)
Answer & Solution
Correct Answer
(A) \((4,2,-4)\)
Step-by-step Solution
Detailed explanation
\(
2 x+y-2 z=18
\)
Dividing both sides \(\sqrt{(2)^2+(1)^2+(-2)^2}=3\), we get
\(
\left(\frac{2}{3}\right) \mathrm{x}+\left(\frac{1}{3}\right) \mathrm{y}-\left(\frac{2}{3}\right) \mathrm{z}=6
\)
Hence length of perpendicular from origin to the plane is 6 . Therefore coordinates of foot of perpendicular are \(\left[\left(\frac{2}{3}\right)(6),\left(\frac{1}{3}\right)(6),\left(\frac{-2}{3}\right)(6)\right]\) i.e. \((4,2,-4)\)
2 x+y-2 z=18
\)
Dividing both sides \(\sqrt{(2)^2+(1)^2+(-2)^2}=3\), we get
\(
\left(\frac{2}{3}\right) \mathrm{x}+\left(\frac{1}{3}\right) \mathrm{y}-\left(\frac{2}{3}\right) \mathrm{z}=6
\)
Hence length of perpendicular from origin to the plane is 6 . Therefore coordinates of foot of perpendicular are \(\left[\left(\frac{2}{3}\right)(6),\left(\frac{1}{3}\right)(6),\left(\frac{-2}{3}\right)(6)\right]\) i.e. \((4,2,-4)\)
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