Find a line through the point po = (-5, 2, 1) and normal to the plane
{(x, y, z): x = y}
The normal vector of the plane is (1, -1, 0).
To find a line through the point po and normal to the plane, we can use the point-normal form of the equation of a line:
r = po + t * n
where r is a point on the line, po is the given point, t is a scalar parameter, and n is the normal vector.
Substituting the values, we have:
r = (-5, 2, 1) + t * (1, -1, 0)
Simplifying, we get:
r = (-5 + t, 2 - t, 1)
Therefore, the line through the point po = (-5, 2, 1) and normal to the plane {(x, y, z): x = y} is given by the equation:
x = -5 + t
y = 2 - t
z = 1
where t is a scalar parameter.
To find a line through a given point and normal to a plane, we can use the following steps:
Step 1: Determine the normal vector of the given plane.
Step 2: Use the given point and the normal vector to create the equation of the line.
Step 1: Determine the normal vector of the given plane:
The equation of the plane given is x = y. To find the normal vector, we need to find the coefficients of x, y, and z. In this case, the coefficients of x and z are 1, and the coefficient of y is -1, since the equation is x - y + 0z = 0. Therefore, the normal vector is (1, -1, 0).
Step 2: Use the given point and the normal vector to create the equation of the line:
We can express the equation of the line as (x, y, z) = (x₀, y₀, z₀) + t(a, b, c), where (x₀, y₀, z₀) is the given point and (a, b, c) is the normal vector.
In this case, the given point is po = (-5, 2, 1) and the normal vector is (1, -1, 0). So, the equation of the line becomes:
(x, y, z) = (-5, 2, 1) + t(1, -1, 0), where t is a parameter.
Therefore, the line through the point po = (-5, 2, 1) and normal to the plane {(x, y, z) : x = y} is:
x = -5 + t
y = 2 - t
z = 1
Please let me know if you need any further assistance.