Find an equation of the plane orthogonal to the line:
(x,y,z) = (5,-6,-1) + t(3,10,-3) which passes through the point (-6,-1,5).
so i got so far:
x=5+3t
y=-6+10t
z=-1-3t
Should be in form ax+by+cz+d=d
Not sure where to go from there...
If the plane is orthogonal to
(5,-6,-1) + t(3,10,-3), then it is also orthogonal to the parallel line
t(3,10,-3) which moves through the origin. Consider first the plane orthogonal to this line which contains the origin. If the point (x,y,z) is on that plane, then:
(x,y,z) dot (3,10,-3) = 0 ---->
3x + 10 y - 3z = 0
The plane we are seeking is parallel to this plane and is therefore given by the equation:
3x + 10 y - 3z = d
You can compute d by demanding that the point (-6,-1,5) lies on the plane.
To find the equation of a plane, you need to find a normal vector to the plane. For a plane orthogonal (perpendicular) to a line, the normal vector will be parallel to the direction vector of the line.
In this case, the direction vector of the line is (3, 10, -3). Let's call it vector d.
The equation of a plane can be written as:
Ax + By + Cz = D
To find A, B, and C, we need a vector that is orthogonal to the plane. Since the plane is orthogonal to the line, we can take the direction vector of the line as a normal vector.
So, A = 3, B = 10, and C = -3.
Now, we can substitute the coordinates of the given point (-6, -1, 5) into the equation to find D:
3x + 10y - 3z = D
Substituting (-6, -1, 5) gives:
3(-6) + 10(-1) - 3(5) = D
-18 - 10 - 15 = D
D = -43
Hence, the equation of the plane orthogonal to the given line and passing through the point (-6, -1, 5) is:
3x + 10y - 3z = -43
To find an equation of the plane orthogonal to the given line, we know that the normal vector of the plane is perpendicular to the given line.
The direction vector of the line is (3, 10, -3). Since a normal vector is perpendicular to any vector in the plane, it must be orthogonal to the direction vector of the line.
Now, to find a normal vector, we can take the dot product of the direction vector of the line and the normal vector and set it equal to zero:
(3, 10, -3) · (a, b, c) = 0
This gives us the equation 3a + 10b - 3c = 0.
Now, we need to find the value of a, b, and c. Since the equation of the plane passes through the point (-6, -1, 5), we can substitute these values into the equation of the plane:
3(-6) + 10(-1) - 3(5) = 0
Simplifying, we get -18 - 10 - 15 = 0, which gives us -43 = 0.
Since -43 is not equal to 0, this means there is no solution for a, b, and c that satisfy the equation.
Therefore, there is no plane orthogonal to the given line that passes through the point (-6, -1, 5).