Find an equation of a plane through the point (1, 5, 1) which is orthogonal to the line

x=3+5t y=5-1t z=-1+4t
in which the coefficient of x is 5.

To find an equation of a plane through the given point that is orthogonal to the line, we need to find the normal vector of the plane and use the point-normal form of the equation of a plane.

First, let's find the direction vector of the line. The direction vector will be the coefficients of t in the equations of x, y, and z.

Direction vector of the line: (5, -1, 4)

Now, since the plane is orthogonal to the line, its normal vector will be perpendicular to the direction vector of the line. We can find the normal vector by taking the cross product of the plane's normal vector and the direction vector.

Normal vector of the plane = (5, -1, 4) x (A, B, C)

Now, we know that the coefficient of x in the equation of the plane is given as 5. So, the normal vector of the plane will be parallel to the vector (5, 0, 0).

Cross product of (5, -1, 4) and (5, 0, 0) gives us:

(5, -1, 4) x (5, 0, 0) = (-1 * 0 - 4 * 0, 4 * 5 - 4 * 0, -1 * 0 - 5 * 0) = (0, 20, 0)

Therefore, the normal vector of the plane is (0, 20, 0).

Now, we can use the point-normal form of the equation of a plane to find an equation of the plane:

Ax + By + Cz = Ax1 + By1 + Cz1

Substituting the given point (1, 5, 1) and the values of A, B, and C derived from the normal vector:

0x + 20y + 0z = 0 * 1 + 20 * 5 + 0 * 1

Simplifying:

20y = 100

Dividing both sides by 20:

y = 5

Therefore, an equation of the plane is:

0x + 20y + 0z = 100

Simplified as:

20y = 100

To find an equation of a plane, we need two things: a point on the plane and a normal vector to the plane. The normal vector of a plane is orthogonal (perpendicular) to every vector lying on the plane.

First, let's find a normal vector to the given line. Since the line is given parametrically, we can find two vectors lying on the line and find their cross product. Let's consider two points on the line, when t = 0 and t = 1:

When t = 0:
x = 3 + 5(0) = 3
y = 5 - 1(0) = 5
z = -1 + 4(0) = -1
Point A: (3, 5, -1)

When t = 1:
x = 3 + 5(1) = 8
y = 5 - 1(1) = 4
z = -1 + 4(1) = 3
Point B: (8, 4, 3)

Now, let's find the vector AB = B - A:
AB = (8 - 3, 4 - 5, 3 - (-1)) = (5, -1, 4)

To find a normal vector to the line, we take the cross product of AB and any other vector on the line. Let's choose the direction vector of the line as the other vector, which is (5, -1, 4):

n = AB x (5, -1, 4)
= (5, -1, 4) x (5, -1, 4)

Calculating the cross product:
n = [(−1)(4) − (−1)(−1), (4)(5) − (5)(4), (5)(−1) − (4)(−1)]
= [0, 0, -9]

So, a normal vector to the line is n = (0, 0, -9).

Now we have a point on the plane (1, 5, 1) and the normal vector to the plane (0, 0, -9). To find an equation of the plane, we can use the point-normal form of the equation of a plane:

ax + by + cz = d

Substituting the point (1, 5, 1) and the normal vector (0, 0, -9):
0(1) + 0(5) + (-9)(1) = d
-9 = d

So, the equation of the plane through the point (1, 5, 1) which is orthogonal to the line x = 3 + 5t, y = 5 - t, z = -1 + 4t, and in which the coefficient of x is 5, is:

5x + 0y + 0z = -9
5x = -9

The direction vector of the line is (5,-1,4)

so a plane perpendicular to that line is
5x - y + 4y + C = 0

but (1,5,1) is on it
5-5+1+C=0
C=-1

5x - y + 4y - 1 = 0