Question

Find a parametric representation for the surface.

The plane that passes through the point (0,-1,5) and contains the vectors\(\left\langle {2,1,4} \right\rangle \) and \(\left\langle { - 3,2,5} \right\rangle \)

Short answer

The parametric representation of the plane is

\(x = 2u - 3v,y = - 1 + u + 2v,z = 5 + 4u + 5v\) where \((u,v) \in {\mathbb{R}^2}\)

Step-by-step solution

Recall and apply

Remember that

The vector equation of plane containing the vectors \({b_1}\) and \({b_2}\)and containing a point with position vector a is \(r(u,r) = a + u{b_1} + v{b_2}\)

Vector equation of plane 

It is given that the plane contains the vectors \(\left\langle { - 3,2,5} \right\rangle \;\) and\(\left\langle {2,1,4} \right\rangle \).

And the plane contains the point with position vector \(\left\langle {0, - 1,5} \right\rangle \)

Therefore, the vector equation of the plane is,

\(\begin{array}{l}r(u,v) = \left\langle {0, - 1,5} \right\rangle + u\left\langle {2,1,4} \right\rangle + v\left\langle { - 3,2,5} \right\rangle \\r(u,v) = \left\langle {0, - 1,5} \right\rangle + \left\langle {2u,1u,4u} \right\rangle + \left\langle { - 3v,2v,5v} \right\rangle \end{array}\).

Therefore, the parametric representation of the plane is

\(x = 2u - 3v,y = - 1 + u + 2v,z = 5 + 4u + 5v\) where \((u,v) \in {\mathbb{R}^2}\).