Question

The co-ordinates of the foot of the perpendicular drawn from the origin to the plane 2x - 3y + 4z = 29 are

• A. (2,3,4)
• B. (2,-3,4)
• C. (2,-3,-4)
• D. (-2,-3,4)

Answer 1

The Correct Option is B

The normal vector of the plane is given by the coefficients of x, y, and z in the equation, which is (2, -3, 4).
This vector is perpendicular to the plane. 
The direction vector of the line perpendicular to the plane is the same as the normal vector of the plane.
 Therefore, the direction vector is (2, -3, 4). 
To find the point of intersection, we can parameterize the line in terms of a parameter
 \(t: x = 2t\ y = -3t\ z = 4t \)
Substituting these values into the equation of the plane, we get: 
\(2(2t) - 3(-3t) + 4(4t)\)
\(=  4t + 9t + 16t \)
\(= 29 t\)
 \(= 1\)
 Therefore, the point of intersection, which is the foot of the perpendicular from the origin to the plane, is: 
\(x = 2(1) = 2 y = -3(1) = -3 z = 4(1) = 4 \)
So, the coordinates of the foot of the perpendicular are\( (2, -3, 4). \)
Therefore, the correct option is (B) (2, -3, 4).

Answer 2

Given: 
Plane equation: \( 2x - 3y + 4z = 29 \)
Origin: \( (0, 0, 0) \)

Step 1: Find the foot of perpendicular from origin to the plane
The formula for foot of perpendicular from point \( (x_0, y_0, z_0) \) to the plane \( Ax + By + Cz + D = 0 \) is:

\[ x = x_0 - A \cdot \frac{Ax_0 + By_0 + Cz_0 + D}{A^2 + B^2 + C^2} \] \[ y = y_0 - B \cdot \frac{Ax_0 + By_0 + Cz_0 + D}{A^2 + B^2 + C^2} \] \[ z = z_0 - C \cdot \frac{Ax_0 + By_0 + Cz_0 + D}{A^2 + B^2 + C^2} \]

Convert plane to general form:
\( 2x - 3y + 4z - 29 = 0 \)
So: \( A = 2, B = -3, C = 4, D = -29 \)

From origin: \( (x_0, y_0, z_0) = (0, 0, 0) \)

Substitute into formula:
\[ \text{Numerator} = 2(0) - 3(0) + 4(0) - 29 = -29 \] \[ \text{Denominator} = 2^2 + (-3)^2 + 4^2 = 4 + 9 + 16 = 29 \] \[ x = 0 - 2 \cdot \frac{-29}{29} = 2 \quad y = 0 - (-3) \cdot \frac{-29}{29} = -3 \quad z = 0 - 4 \cdot \frac{-29}{29} = 4 \]

Answer: \( (2, -3, 4) \)

Answer 3

Given plane: \( 2x - 3y + 4z = 29 \) 

We are to find the foot of the perpendicular from the origin \( (0, 0, 0) \) to this plane. 

Let the foot of the perpendicular be \( (x, y, z) \). Then, the vector from origin to this point is perpendicular to the plane. 

So, it must be parallel to the normal vector of the plane. Normal vector to the plane: \( \vec{n} = \langle 2, -3, 4 \rangle \) 

Let the foot of perpendicular be along the direction of the normal vector: \[ (x, y, z) = \lambda \cdot (2, -3, 4) = (2\lambda, -3\lambda, 4\lambda) \] Now substitute into the plane equation: \[ 2(2\lambda) - 3(-3\lambda) + 4(4\lambda) = 29 \] \[ 4\lambda + 9\lambda + 16\lambda = 29 \] \[ 29\lambda = 29 \Rightarrow \lambda = 1 \] So, foot of the perpendicular is: \[ (2\lambda, -3\lambda, 4\lambda) = (2, -3, 4) \] Answer: (2, -3, 4)