**Geometry and Planes Practice Problems**

If necessary, review:

**PROBLEM 1**

Draw a graph of the plane *W* represented by the following equation:

−2 *x* − 4 *y* + 3 *z* − 12 = 0

**SOLUTION 1**

The *x* -intercept, or the point where the plane *W* intersects the *x* axis, can be found by setting *y* = 0 and *z* = 0 and solving the resulting equation for *x* . Call this point *P* :

Therefore,

*P* = (−6,0,0)

The *y* -intercept, or the point where the plane *W* intersects the *y* axis, can be found by setting *x* = 0 and *z* = 0 and solving the resulting equation for *y* . Call this point *Q* :

Therefore,

*Q* = (0, − 3,0)

The *z* -intercept, or the point where the plane *W* intersects the *z* axis, can be found by setting *x* = 0 and *y* = 0 and solving the resulting equation for *z* . Call this point *R* :

Therefore,

*R* = (0,0,4)

These three points are shown in the plot of Fig. 9-11. The plane can be envisioned, based on this data. (The dashed axes are “behind” the plane.)

**PROBLEM 2**

Suppose a plane contains the point (2,−7,0) and a normal vector to the plane at this point is 3 **i** + 3 **j** + 2 **k** . What is the equation of this plane?

**SOLUTION 2**

The vector 3 **i** + 3 **j** + 2 **k** is equivalent to ( *a,b,c* ) = (3,3,2). We have one point ( *x* _{0} , *y* _{0} , *z* _{0} ) = (2,−7,0). Plugging these values into the general formula for the equation of a plane gives us the following:

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