Determine the value of k so that the line with parametric equations x = 2 + 3t, y = -2 + 5t, z = k is parallel to the plane with equation 4x + 3y – 3z -12 = 0
let k= a+bt
then the line is:
x = 2 + 3t
y = -2 + 5t
z = a + bt and its direction vector is (3,5,b)
If the line is parallel to the plane then it must be perpendicular to the normal of the plane, that is
(4,3,-3)∙(3,5,b) = 0
12 + 15 - 3b = 0
b = 9
so z = a + 9t
No matter what the value of a, the line will be parallel to the plane.
trying to intersect the line with the plane we get
4(2+3t) + 3(-2+5t) - 3(a+9t) = 12
the t's drop out and we get a = -10/3
so if a = -10/3 the line is parallel to but also on the plane, while for any other value of a, the line will not be on the plane, but parallel to it
so possible values of k could be just 9t, or 5+9t, or 1+9t, etc
The vector 4i + 3j - 3k is normal to the given plane. (k is a unit vector here, not the unknown you are looking for, which I will call K). If the line represented by the parametric equations is parallel to the plane, it must be perpendicular to the normal to the plane.
The vector parallel to the line is defined by
t = (x-2)/3 = (y+2)/5 , with no component along the z axis. The line is in the x,y plane regardless of the value of K. Its vector components are
3i + 5j +0k
One can write an equation that requires the two lines to be perpendicular, but it will not involve your unknown K, and there will be no solution, because a line with only x and y components cannot be perpendicular to the plane defined by 4x + 3y – 3z -12 = 0
Are you sure you did not omit a term that involves t in the parametric definition of the line? Is it really z = K ?
Reiny treats your k as an unknown a + bt parametric term, so there are really two unknowns in that case. I treat k as an unknown constant. In Reiny's czse, solutions can be obtained, as he has done.
You should have been clearer about what k is supposed to represent.
I just copied the question as it was...I tried asking a friend and the teacher about k. When I tried to work it out, I assumed that the direction vector was 0, because it was not k+t or any number at all, but I am still unsure. Thanks.
Well, let's take a closer look at the problem. We want to find the value of k that makes the line with parametric equations x = 2 + 3t, y = -2 + 5t, z = k parallel to the plane with equation 4x + 3y – 3z -12 = 0.
To determine if the line is parallel to the plane, we need to check if the direction vector of the line is orthogonal (perpendicular) to the normal vector of the plane.
The direction vector of the line is given by the coefficients of t in the equations: (3, 5, 0).
The normal vector of the plane is the coefficients of x, y, and z in the equation: (4, 3, -3).
Now, if the line is parallel to the plane, then the dot product of the direction vector of the line and the normal vector of the plane should be zero.
Let's calculate the dot product: (3, 5, 0) · (4, 3, -3) = 12 + 15 + 0 - 12 - 9 = 6
Since the dot product is not zero, it means the line is not parallel to the plane for any value of k.
So, in this case, there is no specific value of k that satisfies the condition.
And just like that, the line and the plane could not find a common language to be parallel.
To determine the value of k so that the line with parametric equations x = 2 + 3t, y = -2 + 5t, z = k is parallel to the plane with the equation 4x + 3y – 3z -12 = 0, we need to find the direction vector of the line and the normal vector of the plane. If both vectors are parallel, then the line is parallel to the plane.
First, let's find the direction vector of the line. The direction vector of a line can be obtained by taking the coefficients of t in the parametric equations. In this case, the direction vector is (3, 5, 0).
Next, let's find the normal vector of the plane. The normal vector of a plane can be obtained by taking the coefficients of x, y, and z in the equation of the plane. In this case, the normal vector is (4, 3, -3).
To determine if two vectors are parallel, we can check if their scalar triple product is equal to zero. The scalar triple product of two vectors (a, b, c) and (d, e, f) is given by the determinant:
| i j k |
| a b c |
| d e f | = afj + cdi + bke - (aei + cek + dbf)
For our two vectors (3, 5, 0) and (4, 3, -3), the scalar triple product is:
(0)(3)(3) + (0)(4)(-3) + (5)(4)(-3) - (5)(3)(-3) - (0)(4)(5) - (3)(3)(0) = 0 + 0 - 60 + 45 + 0 - 0 = -15
Since the scalar triple product is not equal to zero, the line is not parallel to the plane.
Therefore, there is no value of k that makes the given line parallel to the plane with equation 4x + 3y – 3z -12 = 0.