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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 = (_,_,_)

Answer :

Answer:

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)

abidemiokin
  • The parametric equation of a line is P(x.y.z)= (-4, 5, -1) + t(-3,0,-2)
  • the coordinate of Q will be at Q (x,y,z) = (-5/2 , 5 , 0)

Parametric equation of a line

The parametric equation of a line is given as:

  • P(x.y.z) = P₀ (x₀,y₀,z₀) + t * v(x,y,z)

where:

  • P₀ is the point of the line at t=0
  • v represents the velocity vector

According to the information, 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 - 2z = 1 hence v = (-3,0,-2)

Get the parametric equation of the line

P(x, y, z) = P₀ (x₀,y₀,z₀) + tv(x,y,z)

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

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

z= z₀ + t*v

z = -1 + (-2)t

t = -1/2

x= x₀ + t* v

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

y= y₀ + t*v

y = 5 + 0*(-1/2) = 5

Hence the coordinate of Q will be at Q (x,y,z) = (-5/2 , 5 , 0)

Learn more on parametric equation of a line here: https://brainly.com/question/8674159

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