Find the parametric equations for the line through the point p = (-4, 5, -1) that is perpendicular to the plane -3x+0y-2z=1. use "t" as your variable, t = 0 should correspond to p, and the velocity vector of the line should be the same as the standard normal vector of the plane. x =y =z =(b) at what point q does this line intersect the yz-plane? q =

for a) P(x.y.z)= (-4, 5, -1) + t*(-3,0,-2)

for b) Q (x,y,z) = (-5/2 , 5 , 0)

Step-by-step explanation:

Since the parametric of the line is given by:

P(x.y.z)=P₀ (x₀,y₀,z₀) + t * v(x,y,z) , where P₀ is the point of the line at t=0 and v represents the velocity vector ( a vector parallel to the line)

since the velocity vector is parallel to the line , and thus perpendicular to the plane, it will be equal to the normal vector of the plane.  In our case:

-3*x + 0*y - 2*z = 1 , thus v=(-3,0,-2)

then

P(x.y.z)=P₀ (x₀,y₀,z₀) + t * v(x,y,z) = (-4, 5, -1) + t*(-3,0,-2)

P(x.y.z)= (-4, 5, -1) + t*(-3,0,-2)

for b)  the line intersects the  x-y plane at z=0 , therefore

z= z₀ + t* vz = -1 + (-2)*t = 0  → t = -1/2

x= x₀ + t* vx = -4 + (-3)*(-1/2) = -5/2

y= y₀ + t* vy = 5 + 0*(-1/2) = 5

Q (x,y,z) = (-5/2 , 5 , 0)

you have orange juice on wednesday. (idk why this is the answer)

step-by-step explanation: