Answer
(a) The equation of the plane in Figure 7(A) is $z=4$.
(b) The equation of the plane in Figure 7(B) is $x=-3$.
(c) The equation of the plane in Figure 7(C) is $x+y=0$.
Work Step by Step
By Eq. (3) of Theorem 1, the equation of a plane in vector form is given by
${\bf{n}}\cdot\left( {x,y,z} \right) = d$,
where ${\bf{n}}$ is the vector normal to the plane.
(a) From Figure 7(A) we see that the plane is parallel to the $xy$-plane, therefore the normal vector is ${\bf{n}} = {\bf{k}}$ or ${\bf{n}} = \left( {0,0,1} \right)$. Thus, the equation of the plane is
$\left( {0,0,1} \right)\cdot\left( {x,y,z} \right) = d$
$z=d$.
Since the plane passes through the point $\left( {0,0,4} \right)$, we obtain $d=4$. Thus, the equation of the plane in Figure 7(A) is $z=4$.
(b) From Figure 7(B) we see that the plane is parallel to the $yz$-plane, therefore the normal vector is ${\bf{n}} = {\bf{i}}$ or ${\bf{n}} = \left( {1,0,0} \right)$. Thus, the equation of the plane is
$\left( {1,0,0} \right)\cdot\left( {x,y,z} \right) = d$
$x = d$.
Since the plane passes through the point $\left( { - 3,0,0} \right)$, we obtain $d=-3$. Thus, the equation of the plane in Figure 7(B) is $x=-3$.
(c) From Figure 7(C) we see that the plane makes an angle $\theta = - \frac{\pi }{4}$ with the positive $x$-axis, therefore the normal vector is ${\bf{n}} = \cos \frac{\pi }{4}{\bf{i}} + \cos \frac{\pi }{4}{\bf{j}} = \frac{1}{2}\sqrt 2 {\bf{i}} + \frac{1}{2}\sqrt 2 {\bf{j}}$ or ${\bf{n}} = \left( {\frac{1}{2}\sqrt 2 ,\frac{1}{2}\sqrt 2 ,0} \right)$. Thus, the equation of the plane is
$\left( {\frac{1}{2}\sqrt 2 ,\frac{1}{2}\sqrt 2 ,0} \right)\cdot\left( {x,y,z} \right) = d$
$\frac{1}{2}\sqrt 2 x + \frac{1}{2}\sqrt 2 y = d$.
Since the plane passes through the origin $\left( {0,0,0} \right)$, we obtain $d=0$. Thus, the equation of the plane in Figure 7(C) is $\frac{1}{2}\sqrt 2 x + \frac{1}{2}\sqrt 2 y = 0$. Simplification yields $x+y=0$.
