Find the value of k so that the line [x,y,z] = [2,-2,0]+ t[2,k,-3] is parallel to the plane kx +2y - 4z= 12
a normal to kx + 2y - 4z = 12 is (k,2,-4)
direction of the given line is (2, k, -3)
if the line is parallel to the plane, it must be perpendicular to the normal
that is,
their dot product = 0
(k,2,-4)o(2,k,-3) = 0
2k + 2k + 12 = 0
4k = -12
k = -3
Thank you!!
yes
To find the value of k so that the line is parallel to the given plane, we need to ensure that the direction vector of the line is orthogonal (perpendicular) to the vector [k, 2, -4] (which is the normal vector of the plane).
Let's first find the direction vector of the line. We are given that the line is represented by the equation [x, y, z] = [2, -2, 0] + t[2, k, -3]. So, the direction vector of the line is [2, k, -3].
Next, we need the dot product of the direction vector of the line and the normal vector of the plane to be zero in order for them to be perpendicular.
The dot product of two vectors, A = [a1, a2, a3] and B = [b1, b2, b3], is given by:
A·B = a1 * b1 + a2 * b2 + a3 * b3.
In this case, we need to find the dot product of the direction vector [2, k, -3] and the normal vector [k, 2, -4] of the plane.
(2 * k) + (k * 2) + (-3 * -4) = 2k + 2k + 12 = 4k + 12.
We want this dot product to be zero, so we have the equation:
4k + 12 = 0.
Solving this equation, we subtract 12 from both sides:
4k = -12.
Then divide both sides by 4:
k = -12/4.
Simplifying the division:
k = -3.
Therefore, the value of k that makes the line parallel to the plane kx + 2y - 4z = 12 is k = -3.