KCET · Maths · Three Dimensional Geometry
The coordinates of foot of the perpendicular drawn from the origin to the plane \(2 x-3 y+4 z=29\) are
- A \((2,3,4)\)
- B \((2,-3,-4)\)
- C \((2,-3,4)\)
- D \((-2,-3,4)\)
Answer & Solution
Correct Answer
(C) \((2,-3,4)\)
Step-by-step Solution
Detailed explanation
\[
\text { Given, equation of plane is } 2 x-3 y+4 z=29
\]
Since, \(O P\) is perpendicular to the plane
\[
2 x-3 y+4 z=29
\]
Therefore, Direction Ratios of \(O P\) will be \(\langle 2,-3,4\rangle\).
Equation of line \(O P\) will be
\[
\frac{x-0}{2}=\frac{y-0}{-3}=\frac{z-0}{4}=\lambda(\text { say })
\]
Coordinates of \(P\) will be \((2 \lambda,-3 \lambda, 4 \lambda)\).
\(\because P\) lies on plane, so \(2(2 \lambda)-3(-3 \lambda)+4(4 \lambda)=29\)
\(\Rightarrow \quad 4 \lambda+9 \lambda+16 \lambda=29\)
\(\Rightarrow \quad \lambda=1\)
\(\therefore\) Coordinates of \(P\) will be \((2,-3,4)\).
\text { Given, equation of plane is } 2 x-3 y+4 z=29
\]
Since, \(O P\) is perpendicular to the plane
\[
2 x-3 y+4 z=29
\]
Therefore, Direction Ratios of \(O P\) will be \(\langle 2,-3,4\rangle\).
Equation of line \(O P\) will be
\[
\frac{x-0}{2}=\frac{y-0}{-3}=\frac{z-0}{4}=\lambda(\text { say })
\]
Coordinates of \(P\) will be \((2 \lambda,-3 \lambda, 4 \lambda)\).
\(\because P\) lies on plane, so \(2(2 \lambda)-3(-3 \lambda)+4(4 \lambda)=29\)
\(\Rightarrow \quad 4 \lambda+9 \lambda+16 \lambda=29\)
\(\Rightarrow \quad \lambda=1\)
\(\therefore\) Coordinates of \(P\) will be \((2,-3,4)\).
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