Consider the Cartesian equation of a line x - 3/2 = y + 1/3 = z - 5/-2 1. Find the vector equation of the line.

Question

Consider the Cartesian equation of a line \(\frac{x-3}{2} = \frac{y+1}{3} = \frac{z-5}{-2}\)

1. Find the vector equation of the line. 
2. Find its intersecting point with the plane 5x + 2y – 6z – 7 = 0 
3. Find the angle made by the line with the plane 5x + 2y – 6z – 7 = 0

Answer 1

1. The vector equation is \(\bar r\) = (3i – j + 5k) + λ(2i + 3 j – 2k).

2. Any point on the line is

\(\frac{x-3}{2} = \frac{y+1}{3} = \frac{z-5}{-2}\) = λ

x = 2λ + 3, y = 3, λ – 1, z = -2λ + 5

Since this lies on the plane ,it satisfies the plane

5(2λ + 3) + 2(3λ – 1) -6(-2λ + 5) – 7 = 0

10λ + 6λ + 12λ + 15 – 2 – 30 – 7 = 0

28λ = 24

λ = \(\frac{6}{7}\)

The point of intersection is  \([\frac{33}{7}, \frac{11}{7}, \frac{23}{7}]\).

3. Let θ be the angle between the line and the plane. The direction of the line and the plane
\(\bar b\) = 2i + 3j + k; \(\bar m\) = 5i + 2j – 6k