Question

Find a basis for the set of vectors in set of real numbers R cubed in the plane x minus 7 y plus 4 z equals 0. ​ [Hint: Think of the equation as a​ "system" of homogeneous​ equations.] A basis for the set of vectors in set of real numbers R cubed in the plane x minus 7 y plus 4 z equals 0 is StartSet nothing EndSet .

Answer 1

Answer:

$\left[\begin{array}{c}7\\1\\0\end{array}\right] and \left[\begin{array}{c}-4\\0\\1\end{array}\right]$

Step-by-step explanation:

Any vector in this plane is actually a solution to the homogeneous system x-7y+4z = 0 (although this system contains only one equation). So we are to find a basis for the kernel of the coefficient matrix

A = x-7y+4z = 0

A =(1, -7, 4)

let x = 7y - 4z

$\left[\begin{array}{c}x\\y\\z\end{array}\right]$   =$\left[\begin{array}{c}7y-4z\\y\\z\end{array}\right] =\left[\begin{array}{c}7y\\y\\0\end{array}\right] +\left[\begin{array}{c}-4z\\0\\z\end{array}\right] = y \left[\begin{array}{c}7\\1\\0\end{array}\right] + z \left[\begin{array}{c}-4\\0\\1\end{array}\right]$

∴$\left[\begin{array}{c}7\\1\\0\end{array}\right] and \left[\begin{array}{c}-4\\0\\1\end{array}\right]$  form a basis for kernel A