1. ## normal vector

Find an equation for the plane that contains the line with parametric equations
x = -7 - 4 t
,
y = -6 + 4 t
,
z = -3 + 8 t
and is parallel to the line with parametric equations
x = 3 t
,
y = -7 + 5 t
,
z = 5 + 8 t
. Find the the normal vector and the EquationOfPlane.

2. Since the lines are both parallel to the plane their cross product is perpindicular to the plane, and therefore is the normal

N = (-4 i + 4j + 8k)x(3i + 5j + 8k)

For a point in the plane use (-7,-6,-3)

3. Originally Posted by Calculus26
Since the lines are both parallel to the plane their cross product is perpindicular to the plane, and therefore is the normal

N = (-4 i + 4j + 8k)x(3i + 5j + 8k)

For a point in the plane use (-7,-6,-3)

Would the equation of the plane be -8*t^2-7,56*t^2-6,-32*t^2-3=0? Because the computer keeps marking this incorrect.

4. The equation should involve x, y, and z --I have no idea what you are doing.

Once you have N = a i + bj + ck

Use a(x -x0) + b(y-y0) + c(z-z0) = 0

to generate the eqn of the plane where (x0,y0,z0) is the known point

5. Originally Posted by Calculus26
The equation should involve x, y, and z --I have no idea what you are doing.

Once you have N = a i + bj + ck

Use a(x -x0) + b(y-y0) + c(z-z0) = 0

to generate the eqn of the plane where (x0,y0,z0) is the known point
I'll tell you what im doing. I took the cross product of these two vectors:

a := [-4*t,4*t,8*t];
b := [3*t,5*t,8*t];

Where do i get the values for the initial and final variable?

6. N = (-4 i + 4j + 8k)x(3i + 5j + 8k)

you don't include the t

N = -8 i +56 j -32 k

you can factor out the -8 and use

N = i -7j + 4k

(x+7) - 7(y+6) + 4(z+3) = 0

I'll let you finish