If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

At the end of this document, after the qa problems (which provide you with questions and solutions), there is a series of Questions, Problems and Exercises.

query_09_4

Question: Find v X w when v = sin(theta)i + cos(theta)j and w = -cos(theta)i + sin(theta)j (theta is any angle).

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Given Solution:

The result is just the vector k:

v X w = ( sin(theta) i + cos(theta) j ) X (-cos(theta) i + sin(theta) j )

= -sin(theta) cos(theta) i X i + sin(theta) sin(theta) i X j - cos(theta) cos(theta) j X i + cos(theta) sin(theta) j X j.

i X i amd j X j are both zero, since sin(theta) = 0 for both of these computations.

i X j = k by the right-hand rule, and likewise j X i . = -k, so the product is

sin(theta) sin(theta) k - cos(theta) cos(theta) (-k) = sin^2(theta) i + cos^2(theta) k = (sin^2(theta) + cos^2(theta) ) * k = k

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Question: Find sin(theta) where theta is the angle between v = -i + j and w = -i + j + 2k.

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Given Solution:

|| v X w || = || v || || w || sin(theta) so

sin(theta) = || v X w || / (|| v || || w || ) = || 2 j + 2 i || / (sqrt(2) sqrt(6) ) = 2 sqrt(2) / ( sqrt(2) sqrt(6) ) = 2 / sqrt(6) = 2 sqrt(6) / 6 = sqrt(6) / 3.

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Question: Find a unit vector which is orthogonal to both v = 2i - j and w = 2j - k.

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Given Solution:

v X w is orthogonal to both v and w.

v X w = i + 2 j +4 k

A unit vector in this direction is

(i + 2 j + 4 k ) / sqrt(1^2 + 2^2 + 4^2) = (i + 2 j + 4 k ) sqrt(21) / 21 .

If we take the dot product of this vector with either of our original vectors we will get zero. You can verify that this vector is indeed a unit vector.

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Question: Find the area of the triangle with vertices P(2,0,0), Q(1,1,-1), R(3,1,2).

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Given Solution:

Consider the vector PQ = < -1, 1, -1 > to be the base of the triangle, which therefore has magnitude || PQ ||.

The vector PR = < 1, 1, 2 > then forms a side adjacent to the base. An altitude from point R to the base then has magnitude || PR || sin(theta).

Since PQ X PR has magnitude || PQ || || PR || sin(theta), which is just the product of the triangle's base and altitude. Thus the area, being have the product of base and altitude, is

triangle area = 1/2 || PQ X PR || .

Having calculated this quantity you will have the area of the triangle.

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Question: 8) Determine if each of the following products is a vector, scalar, or not defined at all. Explain why. u X (v X w) , u dot (v dot w), (u X v) dot (w X r).

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Given Solution:

(v X w) is a vector perpendicular to both v and w , so u X (v X w) is a vector perpendicular to both u and v X w .

(v dot w) is a scalar (i.e., just a number), so u dot (v dot w) is a dot product of a vector with a scalar. Dot products are just defined between vectors, so this expression is not well-defined. That is, this is a meaningless expression.

Both of the cross products (u X v) and (w X r) are vectors, so (u X v) dot (w X r) is a dot product of two vectors, and therefore a scalar.

All these answers assume that none of the vectors is zero, and that none of the cross products are of parallel vectors. In those cases each meaningful calculation would be zero.

All t

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Question: Find a number t such that the vectors -i - j, i - (1/2) j + (1/2)k and -2i -2j - 2tk all lie in the same plane.

Your solution:

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Given Solution:

Any two of these vectors define the orientation of a plane. The direction perpendicular to that plane is perpendicular to all vectors in the plane. If the third vector is also in the plane, it will also be perpendicular to that direction.

Assuming that none of the vectors are zero and that none are parallel to any of the others, we can pick any two of the vectors and find their cross product, which will be perpendicular to the plane. Then the third vector will be in the same plane, provided it is perpendicular to that cross product.

If the vectors are u, v and w, then, our test would be any of the following:

u dot (v X w ) = 0

v dot (u X w ) = 0

w dot (u X v ) = 0.

If any of the vectors is zero, or if any two of the vectors are parallel, then the condition must hold, and you should justify.

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