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I have no idea what to do. Any help would be greatly appreciated.

Take a point P in R3 and two non-zero vectors v and w. Consider the vector equation

Q = P + s(vector v) + t(vector w) for any real s and t:

Let S be the set of all such points Q.

(a) Assume that vector v and vector w are parallel. Show algebraically that S is a line. Sketch of vector v, vector w,

P and the line S.

(b) Assume that vector v and vector w are not parallel.

i. Show that the following points of S are not colinear.

P, P +vector v, P + vector w

ii. For any of the points Q in S, show that

vector PQ is perpendicular to vector n = (vector v cross product vector w).

Remark. 1 Let pi be the plane through P and perpendicular to vector n. Part ii shows that S lies in pi. Part i showed that S is not a line. In fact, S is the plane pi.

iii. Sketch P, vector v, vector w, vector n and the plane S.

(c) Determine whether the following vector equations give you a point, a line or a plane.

i. (x, y, z) = (1, 4, 2) + s< 4,-5, 1 > +t< -8, 10, 3 >

ii. (x, y, z) = (-1, 0, 1) + s< -5, 10, 5 > + t< 3,-6,-3 >

iii. (x, y, z) = (5, 1, 2) + s< 0, 0, 0 > + t< 0, 0, 0 >

(a) If the vectors are parallel, then they are the same line when taken from a fixed point P.

Since this is so, it doesn't matter what multiples of which one are used, for either one puts the points on the same line.

(b) If the vectors v and w are not parallel, then the point could be anywhere in the plane extending from point P in directions v and w.

d

i. If we take P as the origin, v=(1,0,0), and w = (0,1,0), to be co-linear, all combinations of these two vectors should be on a line. Take the three point P, P+v, and P+w. Clearly, if two points are taken to define a line, the third point is not on the line.

ii. It is known that the cross product of two vectors gives the vector that is perpendicular to both. That means any constants could be chosen and a perpendicular vector would be found. This implies that S lies in a plane.

(c) Determine whether the following vector equations give you a point, a line or a plane.

i. (x, y, z) = (1, 4, 2) + s< 4,-5, 1 > + t< -8, 10, 3 >

defines a plane since the two vectors are not quite multiples of each other.

If the vector times s is multiplied by -2, the vector <-8,10,-2> is gotten,

and this is different from <-8,-10,3>.

ii. (x, y, z) = (-1, 0, 1) + s< -5, 10, 5 > + t< 3,-6,-3 > forms a line, since <-5,10,5>

and <3,-6,-3> are both multiples of <1,-2,-1> ( 1st is by -5, 2nd is by 3).

iii. (x, y, z) = (5, 1, 2) + s< 0, 0, 0 > + t< 0, 0, 0 > defines a point since no matter what the values of s and t are, there is nothing added on.

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