Question

The coordinates of the foot of the perpendicular drawn from origin to the plane \(2x - 3y + 4z - 6 = 0\) is :

• A. \(( -\frac{12}{29},\frac{18}{29},\frac{24}{29} )\)
• B. \(( -\frac{12}{29},-\frac{18}{29},-\frac{24}{29} )\)
• C. \(( \frac{12}{29},\frac{18}{29},\frac{24}{29} )\)
• D. \(( \frac{12}{29},-\frac{18}{29},\frac{24}{29} )\)

Answer 1

The Correct Option is D

To determine the coordinates of the foot of the perpendicular drawn from the origin to the plane \(2x - 3y + 4z - 6 = 0\), follow these steps:

1. Let the foot of the perpendicular be point \(P(a, b, c)\).
2. The direction ratios of the normal to the plane are given by the coefficients of \(x\), \(y\), and \(z\) in the plane equation, which are \(2\), \(-3\), and \(4\) respectively.
3. The direction ratios of the line from the origin to \(P(a,b,c)\) must be proportional to these normal direction ratios.
4. Thus, equate each proportionally:
   • \(a = 2k\)
   • \(b = -3k\)
   • \(c = 4k\)
5. Point \(P(a, b, c)\) lies on the plane, so substitute these into the plane equation:
   \(2a - 3b + 4c - 6 = 0\)
   Substituting for \(a\), \(b\), and \(c\):
   \(2(2k) - 3(-3k) + 4(4k) - 6 = 0\)
6. Simplify and solve for \(k\):
   \(4k + 9k + 16k - 6 = 0\)
   \(29k = 6\)
   \(k = \frac{6}{29}\)
7. Now substitute back to find \(a\), \(b\), and \(c\):
   • \(a = 2k = 2\left(\frac{6}{29}\right) = \frac{12}{29}\)
   • \(b = -3k = -3\left(\frac{6}{29}\right) = -\frac{18}{29}\)
   • \(c = 4k = 4\left(\frac{6}{29}\right) = \frac{24}{29}\)

Therefore, the coordinates of the foot of the perpendicular are \(\left( \frac{12}{29},-\frac{18}{29},\frac{24}{29} \right)\).