Question

To find the point of intersection of the parametric equations \(x = 3 - t,y = 2 + t,z = 5t\) with plane equation \(x - y + 2z = 9\).

Short answer

The point of intersection of the parametric equations \(x = 3 - t,y = 2 + t,z = 5t\) with plane equation \(x - y + 2z = 9\) is \((2,3,5)\).

Step-by-step solution

Concept of Parametric Equation

Formula used:

The parametric equations for a line through the point\(\left( {{x_0},{y_0},{z_0}} \right)\)and parallel to the direction vector\(\langle a,b,c\rangle \)are\(x = {x_0} + at,y = {y_0} + bt,z = {z_0} + ct\).

Calculation of the point of intersection

Consider the parametric equations \(x = 3 - t,y = 2 + t,z = 5t\) and the plane equation \(x - y + 2z = 9\).

The scalar parameter is determined by substituting the parametric equations in the equation of the plane.

Substitute \(x = 3 - t,y = 2 + t,z = 5t\) in the plane equation \(x - y + 2z = 9\).

\(\begin{array}{l}(3 - t) - (2 + t) + 2(5t) = 9\\3 - 2 - t - t + 10t = 9\\1 + 8t = 9\\8t = 8\\t = 1\end{array}\)

Substitute \(t = 1\) in the parametric equation \(x = 3 - t,y = 2 + t,z = 5t\) and obtain the coordinate values.

\(\begin{array}{l}x = 3 - 1,y = 2 + 1,z = 5(1)\\x = 2,y = 3,z = 5\end{array}\)

Thus, the point of intersection of the parametric equations \(x = 3 - t,y = 2 + t,z = 5t\) with plane equation \(x - y + 2z = 9\) is \((2,3,5)\).