Let L be the line passing through the point P=(1, −1, −5) with direction vector →d=[−2, −4, 3]T, and let T be the plane defined by −x+y−3z = −1. Find the point Q where L and T intersect.

parametric equations of the line L

x = 1 - 2t
y = -1 - 4t
z = -5+ 3t

plug those values into
-x + y - 3z = -1 and solve for t

once you have t, sub it back into the parametric equations to get the (x,y,z)

Thank you so much

To find the point Q where the line L and the plane T intersect, we need to find the coordinates of Q.

Step 1: Find the equation of the line L.
The equation of a line can be expressed in parametric form using the point-direction vector form: r = P + t * d, where r is a point on the line, P is a point on the line, d is the direction vector, and t is a parameter.

Given P = (1, -1, -5) and d = [-2, -4, 3]ᵀ, the equation of line L is:
r = (1, -1, -5) + t * [-2, -4, 3]ᵀ
= (1 - 2t, -1 - 4t, -5 + 3t)

Step 2: Substitute the equation of line L into the equation of the plane T.
Substituting (1 - 2t, -1 - 4t, -5 + 3t) into the equation of the plane -x + y - 3z = -1, we get:
-(1 - 2t) + (-1 - 4t) - 3(-5 + 3t) = -1

Step 3: Solve for t.
Simplifying the equation, we have:
-1 + 2t - 1 - 4t + 15 - 9t = -1
10 - 11t = -1
10 = 11t - 1
11t = 10 + 1
11t = 11
t = 1

Step 4: Substitute t back into the equation of line L to find the coordinates of Q.
Substituting t = 1 into r = (1 - 2t, -1 - 4t, -5 + 3t), we get:
Q = (1 - 2(1), -1 - 4(1), -5 + 3(1))
= (1 - 2, -1 - 4, -5 + 3)
= (-1, -5, -2)

Therefore, the point Q where the line L and the plane T intersect is Q = (-1, -5, -2).

To find the point Q where line L and plane T intersect, we need to find the coordinates of Q.

Step 1: Find the equation of the line L.
The equation of a line in vector form is given by r = P + t * d, where P is a point on the line, d is the direction vector of the line, and t is a scalar parameter.

In this case, P = (1, -1, -5) and d = [-2, -4, 3]T, so the equation of line L is:
r = (1, -1, -5) + t * [-2, -4, 3]T.

Step 2: Substitute the equation of line L into the equation of plane T to find the point of intersection Q.
The equation of a plane can be written in the form ax + by + cz = d, where (a, b, c) is a normal vector to the plane.

In this case, the equation of plane T is -x + y - 3z = -1. To make it easier to work with, we can rewrite it as x - y + 3z = 1. The normal vector (a, b, c) is therefore (1, -1, 3).

Substituting the equation of line L into the equation of plane T, we get:
(1 - t) - (-1 + 4t) + 3(-5 + 3t) = 1.

Simplifying this equation gives:
1 - t + 1 - 4t - 15 + 9t = 1,
-4t + 9t = 14.

Solving for t, we get:
5t = 14,
t = 14/5.

Step 3: Substitute the value of t back into the equation of line L to find the coordinates of Q.
Substituting t = 14/5 into the equation of line L, we get:
r = (1, -1, -5) + (14/5) * [-2, -4, 3]T.

Calculating this gives:
r = (1, -1, -5) + (14/5) * [-2, -4, 3]T,
= (1, -1, -5) + [-28/5, -56/5, 42/5]T,
= (1 - 28/5, -1 - 56/5, -5 + 42/5)T,
= (-3/5, -9/5, -3/5).

Therefore, the point of intersection Q is Q = (-3/5, -9/5, -3/5).