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Question

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 >

Questioner:Nick
Private:No
Subject:Linear algebra 1

Question:

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 >

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Sorry about the delay.  This was a busy weekend.

Rule: If two vectors are parallel, then either is a scalar multiple of the other.

If (using CAPS for vectors)

Q = P + s V + t W

and if V,W are parallel, then W = r V, so:

Q = P + s V + t r V

Q = P + (s + tr) V

So This would mean that the vector  PQ = (s + tr) V, and so all such vectors PQ are parallel.

But all such vectors CONTAIN point P.  So all must be on the same line.

..........................................

Now if V, W are NOT parallel:

Can P, A = P + V, B = P + W  be collinear (which has two l's)?

Vector PA = V and vector PB = W.

If P, A, and B are collinear, then PA and PB are parallel vectors and thus PA = t PB.

Then V = t W and V,W are parallel, which is a contradiction.

====================================================
ii. For any of the points Q in S, show that
vector PQ is perpendicular to vector n = (vector v cross product vector w).

If Q is in the plane S, then PQ is a vector that is a combination of vectors V, W.  I.e.

PQ = rV + sW.

Now you just have to show that vector N = V cross W is normal to V and also to W

N = V cross W is given by the determinant:

i    j    k
vx  vy   vz
wx  wy   wz

= i( vy wz - vz wy) + j(vz wx - wz vx) + k(vx wy - vy wx)
.......................................
Now dot that with vector V:  vx i  +  vy j  + vz k

= vx vy wz - vx vz wy + vy vz wx - vy wz vx + vz vx wy - vz vy wx

= 0, as you will see after some algebra.

.......................................
And dot that with vector W:  wx i  +  wy j  + wz k ; again you get zero.

So if you dot your normal vector N with any linear combo of V, W, meaning any vector in the plane:

N dot (rV + sW) =

r(N dot V) + s(N dot W) = guess what.

I think you can handle the rest, including part (c)

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