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Find the equations of the planes determined by the given conditions.

The plane contains the point (−4, 1, 3) and is normal to the line determined by

Short Answer

Expert verified

The equation of the plane that contains the point Q=(-4,1,3) and the line determined by r(t)=-4+t,3+5t,2-3t is -x+2y-2z=0

Step by step solution

01

Given information

The plane that contains the point Q=(-4,1,3) and the line determined by r(t)=-4+t,3+5t,2-3t

02

Calculation

The goal is to discover the plane equation that is determined by the given circumstances.

The given point Q=(-4,1,3)is not on the line r(t)=-4+t,3+5t,2-3tas the three different solutions are obtained from the following equation.

-4+t=-43+5t=12-3t=3

The direction vector of the line is d=1,5,-3and the point Q0=(-4,3,2)is on the line r(t)=-4+t,3+5t,2-3t

The vector parallel to the plane of interest is:

Q0Q=-4+4,1-3,3-2=0,-2,1

The normal vector to the plane is the cross product of d=1,5,-3and the vector Q0Q

03

Calculation

The normal vector is:

N=ijk15-30-21

=-1,2,-2

The equation of the plane is:

-1(x+4)+2(y-1)-2(z-3)=0-x+2y-2z-4-2+6=0(Simplify)-x+2y-2z=0(Combine like terms)

The equation of the plane that contains the point Q=(-4,1,3) and the line determined by r(t)=-4+t,3+5t,2-3tis -x+2y-2z=0

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