Calc III help, Finding an equation of the plane?

Question

Having some difficulty with this subject
The plane through the point (-2,8,10) and perpendicular to the line
x=1 +t, y= 2t, z=4-3t

also diff scenario:
Find the point at which the line intersects the given plane:
x=1 + 2t, y =4t, z=2-3t; x+2y-z+1 = 0

Thankyou for your time and help

Answer 1

2nd problem

1+2t + 8t - (2-3t) + 1 = 0
13t = 0
t = 0
(1, 0, 2)

Answer 2

1) Find the equation of the plane through the point
P(-2,8,10) and perpendicular to the line L:

L:
x = 1 + t
y = 2t
z = 4 - 3t

The directional vector v, of line L is:

v = <1, 2, -3>

The directional vector v, of line L is also the normal vector to the plane. With the normal vector v, to the plane and a point on the plane P(-2,8,10), we can write the equation of the plane.

1(x + 2) + 2(y - 8) - 3(z - 10) = 0
x + 2 + 2y - 16 - 3z + 30 = 0
x + 2y - 3z + 16 = 0
_________________

Find the point at which the line L, intersects the given plane.

L:
x= 1 + 2t
y = 4t
z = 2 - 3t

Plane:
x + 2y - z + 1 = 0

Plug in the values of x, y, and z in terms of t and solve for t.

x + 2y - z + 1 = 0
(1 + 2t) + 2(4t) - (2 - 3t) + 1 = 0
1 + 2t + 8t - 2 + 3t + 1 = 0
13t = 0
t = 0

Plug t = 0 into the parametric equation of the line to find the point of intersection.

x= 1 + 2t = 1
y = 4t = 0
z = 2 - 3t = 2

The point of intersection of the line and plane is (1, 0, 2).

Answer 3

putting F(x,y,z) = f(x,y) - z, permits you to evaluate your floor as a point floor. At (a million,2) f(a million,2) = 2 - 4 = -2. So this floor passes by way of (a million, 2, -2). grad(F) = (2xy, x² - 2y, -a million), so at (a million,2, -2), the gradient is (4, -3, -a million). it extremely is the conventional to the plane and (a million, 2, -2) is the ingredient. The tangent plane is 4(x - a million) - 3(y - 2) - (z + 2) = 0 4x - 3y - z = 0 some thing went incorrect along with your working it out. ** replace **** Had an illustration errors. that's fastened now. TY to Jeff getting it top the 1st time!