Determine the co-ordinates of the foot of the perpendicular drawn from the origin to the plane 4x - 2y + 3z - 6 = 0

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  1. \(\frac{12}{\sqrt{45}},-\frac{18}{\sqrt{45}}, \frac{24}{\sqrt{45}}\)
  2. \(\frac{24}{45},-\frac{18}{45}, \frac{12}{45}\)
  3. \(\frac{24}{29},-\frac{12}{29}, \frac{18}{29}\)
  4. \(\frac{18}{\sqrt{29}},-\frac{12}{\sqrt{29}}, \frac{24}{\sqrt{29}}\)

Answer (Detailed Solution Below)

Option 3 : \(\frac{24}{29},-\frac{12}{29}, \frac{18}{29}\)
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Detailed Solution

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Given:

The plane : 4x - 2y + 3z - 6 = 0

Concept:

For a plane having equation ax + by + cz = d(a, b, c) are direction ratios of the normal to the plane.

Equation of a line passing through (x1, y1, z1) and having direction ratios a, b, c in the cartesian form :

\(\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}= \frac{z-z_{1}}{c}\)

Calculation:

Direction ratios of normal to the plane are 4, -2, and 3. 

Equation of a line passing through the origin (0, 0, 0) and having direction ratios 4, -2, 3 is:

\(\frac{x-0}{4}=\frac{y-0}{-2}= \frac{z-0}{3} = λ\)

⇒ x = 4λ , y = -2λ and z = 3λ

Satisfying equation of the plane with above coordinates :

⇒ 4(4λ) - 2(-2λ) + 3(3λ) - 6 = 0

⇒ λ = 6/29

∴ Foot of the perpendicular,

⇒ x = 24/29, y = -12/29 and z = 18/29

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