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Advanced Math/Linear algebra 1


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.
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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Scott A Wilson


I can answer any question in general math, arithetic, discret math, algebra, box problems, geometry, filling a tank with water, trigonometry, pre-calculus, linear algebra, complex mathematics, probability, statistics, and most of anything else that relates to math. I can also say that I broke 5 minutes for a mile, which is over 12 mph, but is that relevant?


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Master of Science at OSU with high honors in mathematics. Bachelor of Science at OSU with high honors in mathematical sciences. This degree involved mathematics, statistics, and computer science. I also took sophmore level physics and chemistry while I was attending college. On the side I took raquetball, but that's still not relevant.

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