Find the point (if it exists) at which the following plane and line intersect. \[ y=-5 ; r(t)=\langle 2 t+2,-t+5, t-6\rangle, \text { for }-\infty

Ask by Ford Tucker. in the United States

Feb 03,2025

To find the intersection point, we can substitute the \(y\) component of the line into the equation of the plane. The plane is defined by \(y = -5\). The parametrization of the line is given by: \[ r(t) = \langle 2t + 2, -t + 5, t - 6\rangle \] We need to find \(t\) such that the \(y\) component of the line, \(-t + 5\), equals \(-5\): \[ -t + 5 = -5 \] Solving this equation: \[ -t = -5 - 5 \implies -t = -10 \implies t = 10 \] Now we can substitute \(t = 10\) back into the line's equation to find the corresponding \(x\) and \(z\): \[ x = 2(10) + 2 = 20 + 2 = 22 \] \[ y = -10 + 5 = -5 \quad \text{(which matches the plane)} \] \[ z = 10 - 6 = 4 \] Thus, the point of intersection of the line and the plane is: \[ (22, -5, 4) \] The point at which the plane and line intersect is \( (22, -5, 4) \).