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Section 12.5: Lines & Planes

Example Problems

EXAMPLE 6 Find the point at which the line with parametric equations \(x = 2 + 3t, y = -4t, z = 5 + t\) intersects the plane \(4x + 5y - 2z = 18\).

SOLUTION We substitute the expressions for \(x, y,\) and \(z\) from the parametric equations into the equation of the plane: \[ 4(2 + 3t) + 5(-4t) - 2(5 + t) = 18 \] This simplifies to \(-10t = 20\), so \(t = -2\). Therefore the point of intersection occurs when the parameter value is \(t = -2\). Then \(x = 2 + 3(-2) = -4, y = -4(-2) = 8, z = 5 - 2 = 3\) and so the point of intersection is \((-4, 8, 3)\).