Determine the value of k so that the line with parametric equations x = 2 + 3t, y = -2 + 5t, z = kt is parallel to the plane with equation 4x + 3y – 3z -12 = 0.
I got the cross product of (2,-2,0) and (4,3,-3) which is (6,6,14), but that's not adding up with the given equation.Where did I mess up?
a line parallel to a plane will be perpendicular to the plane's normal.
So, pick a point in the plane, say (0,0,-4). Two other points might be (0,4,0) and (3,0,0)
That will give you two vectors in the plane,
u = 4j + 4k
and
v = 3i + 4k
Now the normal is w = u×v = 16i + 12j - 12k
Now, find the vector equation of your line. You want k such that
r•w = 0
I see that I made it much too complicated.
You used the cross product, when you should have set the dot product = 0
To determine the value of k that makes the line parallel to the given plane, you need to find the direction vector of the line and check if it is orthogonal (perpendicular) to the normal vector of the plane. If the direction vector is orthogonal to the plane's normal vector, then the line is parallel to the plane.
The direction vector of the line can be obtained by taking the coefficients of t in front of each variable:
Direction vector of the line: <3, 5, k>
The normal vector of the plane can be obtained by taking the coefficients of x, y, and z in the plane equation:
Normal vector of the plane: <4, 3, -3>
To determine if the direction vector is orthogonal to the normal vector, you can take their dot product and check if it equals zero:
<3, 5, k> · <4, 3, -3> = 3(4) + 5(3) + k(-3) = 12 + 15 - 3k
Now, equate the dot product to zero since orthogonal vectors have a dot product of zero:
12 + 15 - 3k = 0
Combine like terms:
27 - 3k = 0
Solve for k:
3k = 27
k = 9
Therefore, the value of k that makes the line parallel to the plane with the equation 4x + 3y – 3z -12 = 0 is k = 9.
It seems like there might have been an error in calculating the dot product. Make sure to double-check your calculations to find the correct value of k.