To find the foot of the perpendicular drawn from point (5,7,3) to the given line, we need to first find a point on the line and the direction vector of the line.
The direction vector of the line is given by the coefficients of x, y, and z in the line equation. Therefore, the direction vector of the line is (3, 8, -5).
Let (a, b, c) be a point on the line. Then, we have:
(a – 15)/3 = (b – 29)/8 = (c – 5)/(-5)
Solving for a, b, and c, we get:
a = 15 + 3t b = 29 + 8t c = 5 – 5t
where t is a parameter.
Now, let’s find the equation of the plane passing through the given point (5,7,3) and perpendicular to the line. The normal vector of the plane is the direction vector of the line. Therefore, the equation of the plane is given by:
3(x – 5) + 8(y – 7) – 5(z – 3) = 0
Simplifying the equation, we get:
3x + 8y – 5z = 94
Now, let’s find the intersection of the line and the plane. Substituting the parametric equations of the line into the equation of the plane, we get:
3(15 + 3t) + 8(29 + 8t) – 5(5 – 5t) = 94
Simplifying the equation, we get:
89t = -6
Therefore, t = -6/89.
Substituting t into the parametric equations of the line, we get:
a = 15 + 3(-6/89) = 129/89 b = 29 + 8(-6/89) = 191/89 c = 5 – 5(-6/89) = 299/89
Therefore, the foot of the perpendicular from point (5,7,3) to the line is (129/89, 191/89, 299/89).
