Assignment 3

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course MTH 277

6/13 between 2-2:30

Question: Find v dot w when v = 4i + j and w =3i + 2k.YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

(4i + j + 0 k) . (3i + 0 j + 2k ) = 4 * 3 + 1 * 0 + 0 * 2 = 12.

confidence rating #$&*:

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Given Solution: v dot w = (4i + j) dot (3i + 2k ) = (4i + j + 0 k) dot (3i + 0 j + 2k ) = 4 * 3 + 1 * 0 + 0 * 2 = 12.

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Self-critique (if necessary):ok

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Self-critique rating:

ok

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Question: Determine whether v = 5i - 5j + 5k and w = 8i - 10j -2k are orthogonal.

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Your solution:

Two vectors are orthogonal if their dot product is zero.

The dot product of these vectors is 5 * 8 - 5 * (-10) + 5 * (-2) = 40 + 50 - 10 = 80.

They are not orthogonal.

confidence rating #$&*:

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3

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Given Solution:

Two vectors are orthogonal if the angle between them is 90 deg, i.e., if and onlye if their dot product is zero.

The dot product of these vectors is 5 * 8 - 5 * (-10) + 5 * (-2) = 40 + 50 - 10 = 80.

They are not orthogonal.

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Self-critique (if necessary):

ok

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Self-critique rating:

ok

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Question: Find the angle between v = 2i +3 k and w = -j + 4k.

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Your solution:

Since v dot w = || v || || w || cos(theta) we have

theta = cos^-1 ( v dot w ) || v || || w || = cos^-1 ( 12 / (sqrt(13) * sqrt( 17) ) = cos^-1 (.79) = 38 degrees.

confidence rating #$&*:

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Given Solution:

Since v dot w = || v || || w || cos(theta) we have

theta = cos^-1 ( v dot w ) || v || || w || = cos^-1 ( 12 / (sqrt(13) * sqrt( 17) ) = cos^-1 (.79) = 38 degrees, approx., or roughly.6 radians.

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Self-critique (if necessary):

ok

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Self-critique rating:

ok

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Question: Find two distinct unit vectors orthogonal to both v = i + 2j -2k and w = i + j - 2k.

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Your solution:

Suppose a i + b j + c k is orthogonal to both.

a + 2 b - 2 c = 0

a + b - 2 c = 0

Subtracting the second equation from the first we get b = 0.

With this value of b both our first and our second equation become

a - 2 c = 0

so that

a = 2 c.

Any vector of the form 2c i + c k is therefore orthogonal to our two vectors.

Any such vector has magnitude sqrt( (2 c)^2 + c^2) = sqrt( 5 c^2) = sqrt(5) | c |.

If c is positive then | c | = c and our vector is

(2 c i + c k ) / (sqrt(5) c) = 2 sqrt(5) / 5 i + sqrt(5) / 5 k.

If c is negative then | c | = - c and our vector will be

(2 c i + c k ) / (- sqrt(5) c) = - 2 sqrt(5) / 5 i - sqrt(5) / 5 k.

Our two solution vectors are equal and opposite. Each is a unit vector perpendicular to the two given vectors.

confidence rating #$&*:

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Given Solution:

Suppose a i + b j + c k is orthogonal to both. Then the dot product of this vector with each of the given vectors is zero, and we have

a + 2 b - 2 c = 0

a + b - 2 c = 0

Subtracting the second equation from the first we get b = 0.

With this value of b both our first and our second equation become

a - 2 c = 0

so that

a = 2 c.

Any vector of the form 2c i + c k is therefore orthogonal to our two vectors.

Any such vector has magnitude sqrt( (2 c)^2 + c^2) = sqrt( 5 c^2) = sqrt(5) | c |.

If c is positive then | c | = c and our vector is

(2 c i + c k ) / (sqrt(5) c) = 2 sqrt(5) / 5 i + sqrt(5) / 5 k.

If c is negative then | c | = - c and our vector will be

(2 c i + c k ) / (- sqrt(5) c) = - 2 sqrt(5) / 5 i - sqrt(5) / 5 k.

Our two solution vectors are equal and opposite. Each is a unit vector perpendicular to the two given vectors.

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Self-critique (if necessary):

ok

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Self-critique rating:

ok

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Question: Let v = i - j + 4k and w = -i + 3j + 2k. Find cos(theta). Find s such that v is orthogonal to sv - w. Also find t such that v - tw is orthogonal to w

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Your solution:

cos(theta) = v dot w / ( || v || || w ||) = 4 / (sqrt(18) sqrt(14) ) = 4 / (12 sqrt(7) ).

The condition v orthogonal to s v - w is

v dot (s v - w ) = 0

(i - j + 4 k ) dot ( (s - 1) i + (-s + 3) j + (4 s + 2) k ) = 0

which becomes

s - 1 + s - 3 + 16 s + 8 = 0

so that

18 s = 4

and

s = 4 / 18 = 2/9.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3

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Given Solution:

cos(theta) = v dot w / ( || v || || w ||) = 4 / (sqrt(18) sqrt(14) ) = 4 / (12 sqrt(7) ).

The condition v orthogonal to s v - w is

v dot (s v - w ) = 0

(i - j + 4 k ) dot ( (s - 1) i + (-s + 3) j + (4 s + 2) k ) = 0

which becomes

s - 1 + s - 3 + 16 s + 8 = 0

so that

18 s = 4

and

s = 4 / 18 = 2/9.

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Self-critique (if necessary):ok

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Self-critique rating:ok

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Question: Find the work performed when a force F = (6/11)i - (2/11)j + (6/11)k is applied to an object moving along the line from P(3,5,-4) to Q(-4,-9,-11).

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Your solution:

The work is F dot `ds = ( (6/11)i - (2 / 11) j + (6 / 11) k ) dot (-7 i - 14 j - 7 k ) = -42/11 + 28 / 11 - 42 /11 = -56 / 11.

confidence rating #$&*:

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Given Solution: The work is F dot `ds = ( (6/11)i - (2 / 11) j + (6 / 11) k ) dot (-7 i - 14 j - 7 k ) = -42/11 + 28 / 11 - 42 /11 = -56 / 11.

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Self-critique (if necessary):

ok

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Self-critique rating:

ok

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Self-critique (if necessary):

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Self-critique rating:

ok

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Self-critique (if necessary):

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Self-critique rating:

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