Given two planes in 3D space, Plane 1: \(3x - 4y + 2z = 12\) and Plane 2: \(x + 2y - z = 5\). Find the intersection of the line that passes through the origin and is perpendicular to Plane 1 with Plane 2.
Answer
Step 4: Substitute \(t = -\frac{5}{3}\) back into the equations of the line to get the intersection point. \(x = 3t = -5\), \(y = -4t = \frac{20}{3}\), \(z = 2t = -\frac{10}{3}\).
Step 1 ：Step 1: Find the normal vector of Plane 1. Since the coefficients of \(x\), \(y\), and \(z\) in the equation of a plane give the normal vector, the normal vector of Plane 1 is \(\vec{n}_1 = (3, -4, 2)\).
Step 2 ：Step 2: Since the line is perpendicular to Plane 1 and passes through the origin, its parametric equations are \(x = 3t\), \(y = -4t\), and \(z = 2t\).
Step 3 ：Step 3: Substitute the equations of the line into the equation of Plane 2 to find the intersection point. \(3t + 2(-4t) - 2t = 5\) => \(-3t = 5\) => \(t = -\frac{5}{3}\).
Step 4 ：Step 4: Substitute \(t = -\frac{5}{3}\) back into the equations of the line to get the intersection point. \(x = 3t = -5\), \(y = -4t = \frac{20}{3}\), \(z = 2t = -\frac{10}{3}\).
