## Linear Algebra and Its Applications, Exercise 3.1.21

Exercise 3.1.21. If $P$ is the plane in $\mathbb{R}^3$ described by $x+2y-z = 6$ what is the equation for the plane $P'$ parallel to $P$ through the origin? What is a vector perpendicular to $P'$? Find a matrix $A$ for which $P'$ is the nullspace, and a matrix $B$ for which $P'$ is the row space.

Answer:  One way to approach this problem is to find a general solution to the equation $x+2y-z = 6$ and express it as the sum of a homogeneous solution and a particular solution. The corresponding homogeneous system is $x+2y-z = 0$, which is represented by the matrix $A = \begin{bmatrix} 1&2&-1 \end{bmatrix}$

This system has $x$ as a basic variable and $y$ and $y$ as free variables. Setting $y = 1$ and $z = 0$ we have $x+2y-z = x+2-0 = 0$ or $x = -2$. So $(-2, 1, 0)$ is one solution to the homogeneous system. Setting $y = 0$ and $z = 1$ we have $x+2y-z = x+0-1 = 0$ or $x = 1$. So $(1, 0, 1)$ is a second solution to the homogeneous system. These vectors are in the nullspace of $A$ and serve as a basis for it.

To find the particular solution we set $y = z = 0$ for the general system so that we have $x+2y-z = x+0-0 = 6$ or $x = 6$. So $(6, 0, 0)$ is a particular solution to the general system, and the general solution to the equation is then the sum of the particular solution and the homogeneous solution: $\begin{bmatrix} 6 \\ 0 \\ 0 \end{bmatrix} + y \begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix} + z \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}$

The plane $P$ is defined by the general solution to $x+2y-z=6$. The plane $P'$ going through the origin corresponds to the homogeneous system $x+2y-z=0$ and is spanned by the vectors $(-2, 1, 0)$ and $(1, 0, 1)$; it is the nullspace of the matrix $A$ above. The plane $P$ is parallel to $P'$ and is offset from it by the vector $(6, 0, 0)$.

To find a vector perpendicular to $P'$ (and $P$),  from the above solution to the homogeneous system $x+2y-z = 0$ we know that the vector $(1, 2, -1)$ (the first and only row of the matrix $A$ above) is orthogonal to the vectors $(-2, 1, 0)$ and $(1, 0, 1)$ (the basis vectors for the nullspace of $A$). Therefore the vector $(1, 2, -1)$ is perpendicular to the plane $P'$, the nullspace of $A$ spanned by $(-2, 1, 0)$ and $(1, 0, 1)$.

Using the vectors $(-2, 1, 0)$ and $(1, 0, 1)$ that serve as a basis for the plane $P'$ we can construct a matrix $B$ for which $P'$ is the row space: $B = \begin{bmatrix} -2&1&0 \\ 1&0&1 \end{bmatrix}$

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition , Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books .

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