Answer:
a) (x, y, z) = (1, -1, 0)
b) a = 0; L = {t, t, 0}
c) a = 1 . . or . . a = 3
d) No
Step-by-step explanation:
a) The solution using row-reduction techniques is shown in the first attachment, part (a). The generic solution is shown evaluated for a=2, the result being (1, -1, 0).
You will note that the solution gives indeterminate values for a=1 and a=3, which make the denominators zero.
b) For this, we use the point solution of part (a) together with the cross product of the normal vectors of planes 1 and 2. Said cross product will be in the direction of their line of intersection. We substitute the parametric equation for that line into the equation for plane 3 and solve for <em>a</em>. The result is a = 0.
The simplification of the resulting equation also gives a=1 and a=3, but we treat those as extraneous solutions because the corresponding common point of intersection of the planes does not exist.
c) The values a = 1 and a = 3 were found in part (a). These are values of <em>a</em> that make the common point indeterminate.
d) Substituting the given values for (x, y, z) gives three equations for <em>a</em> that have 3 different solutions. Hence there cannot be a value of <em>a</em> that makes this be a common point of intersection.
e) The second attachment illustrates part (a), with <em>a</em>=2. The third attachment illustrates part (b) with <em>a</em>=0.
The fourth attachment illustrates part (c). The first picture has the triangular tunnel between the planes in a vertical orientation (<em>a</em>=1). The second picture has the triangular tunnel extending from lower front to upper back (<em>a</em>=3).
