Find the Intersection of the Line Perpendicular to Plane 1 Through the Origin and Plane 2 y=3x+2 , x-4y=9

Problem

Plane 1: *y=3*x+2, Plane 2: *x−4*y=9

Solution

1. Identify the normal vector of Plane 1. The equation y=3*x+2 can be rewritten in general form as 3*x−y+0*z+2=0 The normal vector n is the vector of coefficients of x y and z

n=<3,−1,0>

2. Determine the equation of the line. The line is perpendicular to Plane 1, so its direction vector is n=<3,−1,0> Since it passes through the origin (0,0,0) the parametric equations are:

x=3*t

y=−t

z=0

3. Substitute the parametric equations of the line into the equation for Plane 2, x−4*y=9 to find the value of the parameter t at the intersection point.

(3*t)−4*(−t)=9

4. Solve for t by combining like terms.

3*t+4*t=9

7*t=9

t=9/7

5. Calculate the coordinates of the intersection point by substituting t=9/7 back into the parametric equations of the line.

x=3*(9/7)=27/7

y=−(9/7)=−9/7

z=0

Final Answer

(x,y,z)=(27/7,−9/7,0)

Want more problems? Check here!