Question

Find a set of parametric equations of the line. The line passes through the point (2,3,4) and is perpendicular to the plane given by \(3 x+2 y-z=6\).

Short answer

The parametric equations of the line that is perpendicular to the plane \(3x + 2y - z = 6\) and passes through the point (2,3,4) are: \(x = 2 + 3t\), \(y = 3 + 2t\), \(z = 4 - t\).

Step-by-step solution

Identifying the Normal Vector

The normal vector \(N\) to the plane given by the plane equation \(3x + 2y - z = 6\) is formed by the coefficients of \(x\), \(y\), and \(z\). Therefore, the normal vector \(N\) is \(N=(3, 2, -1)\).

Formulating the Parametric Equations

The line we are looking for is perpendicular to the plane, therefore, it is parallel to the normal vector of the plane. This means we can use the normal vector to form the direction ratios of the parametric equations. As the line also passes through the point (2, 3, 4), we can use this information to form the constant term in the parametric equations. The parametric equations are: \(x = 2 + 3t\),\(y = 3 + 2t\),\(z = 4 - t\). Here, \(t\) is the parameter.