Trigonometry MathHands.com Sec. 06 exercise M´arquez Euler...

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Euler’s Identity

1. Convert Convert 3ei30◦

to standard form, a + bi

Solution: According to Euler’s amazing identity..eiθ = cos θ + i sin θ..thus..

3ei30◦

= 3[cos 30◦ + i sin 30◦]

Alternatively, you could convert using the picture... and convert in very much the same way we convert frompolar to cartesian...

“√

2. Convert Convert 3e−i30◦

to standard form, a + bi

3e−i30◦

= 3[cos−30◦ + i sin−30◦]

Or use the picture... Or use polar conversion ideas.. (r, θ) = (3,−30◦)

3. Convert Convert ei π

2 to standard form, a + bi

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2 = cosπ

2+ i sin

2= 0 + i(1)

4. Convert Convert e3π

2i to standard form, a + bi

5. Convert Convert e7π

6i to standard form, a + bi

6. Convert Convert 3e2i to standard form, a + bi

7. Convert Convert 5e3.14i to standard form, a + bi

8. Convert Convert e6.28i to standard form, a + bi

9. Convert Convert 2 + i to Euler form, reiθ

Solution: 2 + i =√

5ei26.57◦

many other correct possibilities for theta..

10. Convert Convert 2 − 2i to Euler form, reiθ

Solution: 2 − 2i = (2√

2)e−i45◦

OR (2√

2)ei315◦

or (2√

2)ei765◦

or infinite many more possibilities for θ

11. Convert Convert −2 − 2i to Euler form, reiθ

Solution: −2 − 2i = (2√

2)ei225◦

many other correct possibilities for theta..

12. Convert Convert −3 + 2i to Euler form, reiθ

13. Convert Convert i to Euler form, reiθ

Solution: i = ei90◦

many other correct possibilities for theta.. note r = 1

14. Convert Convert 3i to Euler form, reiθ

Solution: i = 3ei90◦

15. Convert Convert .5 + 3i to Euler form, reiθ

16. Convert Convert −1 to Euler form, reiθ

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17. Convert Convert −i to Euler form, reiθ

18. Euler’s World Multiply or divide as indicated

(a) Multiply the old fashion way...

(b) Multiply the Euler way, and simplify answer.. (note

these are the same numbers..)

e30◦i · e60

Solution: ei90◦

3e30◦i · 2e60

Solution: 6ei90◦

·(√

e30◦i · e30

Solution: ei60◦

(a) Divide the old fashion way...

(b) Divide the Euler way, and simplify answer.. (note

e30◦i

e60◦i

Solution: e−i30◦

(a) Expand the binomial, the old fashion way...

(b) Calculate, the Euler way, and simplify answer.. (notethese are the same numbers..)

e30◦i)3

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Solution: ei90◦

= i compare this problem with#57 from two sections ago.

(a) Expand the binomial, the old fashion way (or not..)...

(b) Calculate, the Euler way, and simplify answer.. (note

e30◦i)7

Solution: ei210◦

24. Euler’s World Convert the numbers to Euler form, then compute.

(3 + 4i)(−2 − 5i)

(b)3 + 4i

−2 − 5i

25. Euler’s World to find Roots

(a) Try to find a square root of i, i.e. a number x suchthat x2 = i

(b) Find a root, the Euler way, and simplify answer..

(note these are the same numbers..)

e90◦i)

Solution: ei45◦

(a) Try to find a square root: (after a good honest at-tempt, go to part b.)

(b) Find a root, the Euler way, and simplify answer..

(note these are the same numbers..)

e60◦i)

Solution: ei30◦

(a) Try to find a fourth root: (after a good honest at-tempt, go to part b.)

(b) Find a fourth root, the Euler way, and simplify an-

swer.. (note these are the same numbers..)

16e60◦i)

Solution: 2ei15◦

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28. FTA and Unity Roots The two solutions to x2 = 1 are x = 1,−1. Show how one can derive these solutions.

29. FTA and Unity Roots The three solutions to x3 = 1 are x = 1, −1

2i, −1

2i. Show how one can derive

these solutions.

30. FTA and Unity Roots The four solutions to x4 = 1 are x = 1, i,−1,−i. Show how one can derive these solutions.

31. FTA and Unity Roots To find the five solutions to x5 = 1, start with writing 1 = ek360◦i, then start finding 5th

roots, every value of k should produce one...

Solution: for integers k, eik72◦

32. Euler’s World Another Proof of MOTA

(a) Multiply the old fashion way

(cosx + i sinx)(cos y + i sin y)

(b) multiply, the Euler way, and simplify answer.. (notethese are the same numbers..)

eix · eiy

(c) Compare the real part of the answer for a, with thereal part of the answer for part b.

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1. Vector Arithmetic

(a) Vector ADDITION Suppose ~v = <1, 2> and ~w = < − 3, 5>, compute

~v + ~w

Solution:

~v + ~w = <1, 2> + < − 3, 5> (given)

= <1 + −3, 2 + 5> (definition of addition on vectors)

= < − 2, 7> (by inspection)

(b) Vector ADDITION Suppose ~v = <1, 3> and ~w = <1,−6>, compute

~v + ~w

Solution:

~v + ~w = <1, 3> + <1,−6> (given)

= <1 + 1, 3 + −6> (definition of addition on vectors)

= <2, − 3> (by inspection)

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(c) Vector ADDITION Suppose ~v = <1, 1> and ~w = <2,−3>, compute

~v + ~w

Solution:

~v + ~w = <1, 1> + <2,−3> (given)

(d) Vector ADDITION Suppose ~v = <5, 2> and ~w = < − 3, 1>, compute

~v + ~w

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Solution:

~v + ~w = <5, 2> + < − 3, 1> (given)

= <5 + −3, 2 + 1> (definition of addition on vectors)

= <2, 3> (by inspection)

(e) Vector ADDITION Suppose ~v = <5, 1> and ~w = <1,−2>, compute

~v + ~w

Solution:

~v + ~w = <5, 1> + <1,−2> (given)

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~v + ~w

(f) Vector ADDITION Suppose ~v = <6, 2> and ~w = <2, 3>, compute

~v + ~w

Solution:

~v + ~w = <6, 2> + <2, 3> (given)

= <6 + 2, 2 + 3> (definition of addition on vectors)

(g) Vector ADDITION Suppose ~v = <5, 1> and ~w = <1,−2>, compute

~v + ~w

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Solution:

~v + ~w = <5, 1> + <1,−2> (given)

~v + ~w

(h) Vector ADDITION Suppose ~v = < − 3, 2> and ~w = < − 1, 5>, compute

~v + ~w

Solution:

~v + ~w = < − 3, 2> + < − 1, 5> (given)

= < − 3 + −1, 2 + 5> (definition of addition on vectors)

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(a) Vector SUBTRACTION Suppose ~v = <1, 2> and ~w = < − 3, 5>, compute

~v − ~w

Solution:

~v − ~w = <1, 2> − < − 3, 5> (given)

= <1 −−3, 2 − 5> (definition of subtraction on vectors)

= <4,−3> (by inspection)

(b) Vector SUBTRACTION Suppose ~v = <1, 3> and ~w = <1,−6>, compute

~v − ~w

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Solution:

~v − ~w = <1, 3> − <1,−6> (given)

= <1 − 1, 3 −−6> (definition of subtraction on vectors)

~w~v−

(c) Vector SUBTRACTION Suppose ~v = <1, 1> and ~w = <2,−3>, compute

~v − ~w

Solution:

~v − ~w = <1, 1> − <2,−3> (given)

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(d) Vector SUBTRACTION Suppose ~v = <5, 2> and ~w = < − 3, 1>, compute

~v − ~w

Solution:

~v − ~w = <5, 2> − < − 3, 1> (given)

= <5 −−3, 2 − 1> (definition of subtraction on vectors)

~v − ~w

(e) Vector SUBTRACTION Suppose ~v = <5, 1> and ~w = <1,−2>, compute

~v − ~w

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Solution:

~v − ~w = <5, 1> − <1,−2> (given)

(f) Vector SUBTRACTION Suppose ~v = <6, 2> and ~w = <2, 3>, compute

~v − ~w

Solution:

~v − ~w = <6, 2> − <2, 3> (given)

= <6 − 2, 2 − 3> (definition of subtraction on vectors)

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~v − ~w

(g) Vector SUBTRACTION Suppose ~v = <5, 1> and ~w = <1,−2>, compute

~v − ~w

Solution:

~v − ~w = <5, 1> − <1,−2> (given)

(h) Vector SUBTRACTION Suppose ~v = < − 3, 2> and ~w = < − 1, 5>, compute

~v − ~w

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Solution:

~v − ~w = < − 3, 2> − < − 1, 5> (given)

= < − 3 −−1, 2 − 5> (definition of subtraction on vectors)

= < − 2,−3> (by inspection)

−~w ~v−

(a) Vector SCALARS Suppose ~v = <2, 0>compute

Solution:

2~v = 2<2, 0> (given)

= <2(2), 2(0)> (def of scalar multiplication)

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(b) Vector SCALARS Suppose ~v = <2, 0>compute

Solution:

3~v = 3<2, 0> (given)

(c) Vector SCALARS Suppose ~v = <2, 0>compute

−2~v

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Solution:

−2~v = −2<2, 0> (given)

= < − 2(2),−2(0)> (def of scalar multiplication)

−2~v

(d) Vector SCALARS Suppose ~v = <4, 2>compute

Solution:

2~v = 2<4, 2> (given)

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(e) Vector SCALARS Suppose ~v = <4,−2>compute

Solution:

3~v = 3<4,−2> (given)

= <3(4), 3(−2)> (def of scalar multiplication)

~v 3~v

(f) Vector SCALARS Suppose ~v = <1, 3>compute

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Solution:

2~v = 2<1, 3> (given)

(g) Vector SCALARS Suppose ~v = <1, 3>compute

Solution:

3~v = 3<1, 3> (given)

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(h) Vector SCALARS Suppose ~v = <1, 3>compute

Solution:

1.5~v = 1.5<1, 3> (given)

= <1.5(1), 1.5(3)> (def of scalar multiplication)

= <1.5, 4.5> (by inspection)

(i) Vector SCALARS Suppose ~v = <1, 3>compute

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Solution:

.75~v = .75<1, 3> (given)

= <.75(1), .75(3)> (def of scalar multiplication)

= <0.75, 2.25> (by inspection)

(j) Vector SCALARS Suppose ~v = <4, 2>compute

−1~v

Solution:

−1~v = −1<4, 2> (given)

= < − 1(4),−1(2)> (def of scalar multiplication)

= < − 4,−2> (by inspection)

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−1~v

4. Vector Arithmetic: Famous Vectors There are two very famous vectors, there are: i =< 1, 0 > and j =< 0, 1 >.

(a) FAMOUS VECTORS i and j Compute and draw the following vectors

4i + 2j

Solution: let

~v = 4i + 2j

= 4 < 1, 0 > +2 < 0, 1 > (given)

=< 4, 0 > + < 0, 2 > (def of scalar multiplication)

=< 4, 2 >

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(b) FAMOUS VECTORS i and j Compute and draw the following vectors

1i + −2j

Solution: let

~v = 1i + −2j

= 1 < 1, 0 > + − 2 < 0, 1 > (given)

=< 1, 0 > + < 0,−2 > (def of scalar multiplication)

=< 1,−2 >

1i+−

(c) FAMOUS VECTORS i and j Compute and draw the following vectors

5i + −2j

Solution: let

~v = 5i + −2j

= 5 < 1, 0 > + − 2 < 0, 1 > (given)

=< 5, 0 > + < 0,−2 > (def of scalar multiplication)

=< 5,−2 >

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5i +−2j

(d) FAMOUS VECTORS i and j Compute and draw the following vectors

5i + 1j

Solution: let

~v = 5i + 1j

= 5 < 1, 0 > +1 < 0, 1 > (given)

=< 5, 1 >

5i + 1j

(e) FAMOUS VECTORS i and j Compute and draw the following vectors

6i + 3j

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Solution: let

~v = 6i + 3j

= 6 < 1, 0 > +3 < 0, 1 > (given)

=< 6, 3 >

(f) FAMOUS VECTORS i and j Compute and draw the following vectors

3i + 4j

Solution: let

~v = 3i + 4j

= 3 < 1, 0 > +4 < 0, 1 > (given)

=< 3, 4 >

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5. NORM of a VECTOR Find the norm of the indicated vector.

(a) Compute:

|| < 4, 2 > ||

Solution:

|| < 4, 2 > || =√

(4)2 + (2)2 (def of norm)

20 (by Inspection)

≈ 4.47 (by Calc)

≈4.4

(b) Compute:

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|| < 1,−2 > ||

Solution:

|| < 1,−2 > || =√

(1)2 + (−2)2 (def of norm)

5 (by Inspection)

≈ 2.24 (by Calc)

≈2.2

(c) Compute:

|| < 5,−2 > ||

Solution:

|| < 5,−2 > || =√

(5)2 + (−2)2 (def of norm)

29 (by Inspection)

≈ 5.39 (by Calc)

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≈5.39

(d) Compute:

|| < 5, 1 > ||

Solution:

|| < 5, 1 > || =√

(5)2 + (1)2 (def of norm)

26 (by Inspection)

≈ 5.1 (by Calc)

≈ 5.1

(e) Compute:

|| < 6, 3 > ||

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Solution:

|| < 6, 3 > || =√

(6)2 + (3)2 (def of norm)

45 (by Inspection)

≈ 6.71 (by Calc)

≈6.7

(f) Compute:

|| < 5, 0 > ||

Solution:

|| < 5, 0 > || =√

(5)2 + (0)2 (def of norm)

25 (by Inspection)

≈ 5 (by Calc)

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5 0≈ 5

6. Which is Bigger?

(a) Determine which number is larger: ||4i + 2j|| OR (||4i|| + ||2j||)

Solution:

||4i + 2j|| = || < 4, 2 > || (def of i and j)

(4)2 + (2)2 (def of norm)

20 (by Inspection)

≈ 4.47 (by Calc)

Meanwhile

||4i|| + ||2j|| = || < 4, 0 > || + || < 0, 2 > || (def of i and j)

= 4 + 2 (def of norm)

= 6 (by Inspection)

Therefore, ||4i|| + ||2j|| is larger.

(b) Determine which number is larger: ||1i + −2j|| OR (||1i|| + || − 2j||)

Solution:

||1i + −2j|| = || < 1,−2 > || (def of i and j)

(1)2 + (−2)2 (def of norm)

5 (by Inspection)

≈ 2.24 (by Calc)

Meanwhile

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||1i|| + || − 2j|| = || < 1, 0 > || + || < 0,−2 > || (def of i and j)

= 3 (by Inspection)

Therefore, ||1i|| + || − 2j|| is larger.

(c) Determine which number is larger: ||5i + −2j|| OR (||5i|| + || − 2j||)

Solution:

||5i + −2j|| = || < 5,−2 > || (def of i and j)

(5)2 + (−2)2 (def of norm)

29 (by Inspection)

≈ 5.39 (by Calc)

Meanwhile

||5i|| + || − 2j|| = || < 5, 0 > || + || < 0,−2 > || (def of i and j)

= 7 (by Inspection)

Therefore, ||5i|| + || − 2j|| is larger.

(d) Determine which number is larger: ||5i + 1j|| OR (||5i|| + ||1j||)

Solution:

||5i + 1j|| = || < 5, 1 > || (def of i and j)

(5)2 + (1)2 (def of norm)

26 (by Inspection)

≈ 5.1 (by Calc)

Meanwhile

||5i|| + ||1j|| = || < 5, 0 > || + || < 0, 1 > || (def of i and j)

= 6 (by Inspection)

(e) Determine which number is larger: ||6i + 3j|| OR (||6i|| + ||3j||)

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Solution:

||6i + 3j|| = || < 6, 3 > || (def of i and j)

(6)2 + (3)2 (def of norm)

45 (by Inspection)

≈ 6.71 (by Calc)

Meanwhile

||6i|| + ||3j|| = || < 6, 0 > || + || < 0, 3 > || (def of i and j)

= 9 (by Inspection)

(f) Determine which number is larger: ||5i + 0j|| OR (||5i|| + ||0j||)

Solution:

||5i + 0j|| = || < 5, 0 > || (def of i and j)

(5)2 + (0)2 (def of norm)

25 (by Inspection)

≈ 5 (by Calc)

Meanwhile

||5i|| + ||0j|| = || < 5, 0 > || + || < 0, 0 > || (def of i and j)

= 5 (by Inspection)

(g) ||u + v|| OR (||u|| + ||v||)

Solution: the idea is to study the above pattern and understand that it will always be the case that thesum of the individual norms is larger or equal to the norm of the sum of the vectors. In some sense, this isequivalent to saying that the sum of the lengths of any two sides of a [Euclidean] triangle have to be largerthan the size of the length of the the third side of the triangle. This is very famous, it is called the triangleinequality.

7. Normalize this.. Find the normalized vector for each:

(a) NORMALIZE them Compute and draw the corresponding normalized vector:

~v =< 5, 2 >

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Solution: First we find the norm of the vector:

||~v|| = || < 5, 2 > || =√

(5)2 + (2)2 (def of norm)

29 (by Inspection)

≈ 5.39 (by Calc)

Then we scale the original vector ~v by multiplying by 1

||~v|| . Let us denote the normalized vector ~v as ” ~vn”

~vn =1

||~v||<5, 2> (given)

||~v|| ,2

||~v||

(def of scalar multiplication)

≈⟨

(approximate)

≈ 〈0.93, 0.37〉 (by inspection)

Now, NOTE: ~v and ~vn have the same direction, BUT, ~vn has norm ”1” as intended. (also keep in mindposition does not matter for vectors.. only direction and size, thus nothing should be interpreted from theirposition, only their size and direction)

(b) NORMALIZE them Compute and draw the corresponding normalized vector:

~v =< 5,−2 >

||~v|| = || < 5,−2 > || =√

(5)2 + (−2)2 (def of norm)

29 (by Inspection)

≈ 5.39 (by Calc)

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~vn =1

||~v||<5,−2> (given)

||~v|| ,−2

||~v||

≈⟨

5.39,−2

(approximate)

≈ 〈0.93,−0.37〉 (by inspection)

(c) NORMALIZE them Compute and draw the corresponding normalized vector:

~v =< −3, 4 >

||~v|| = || < −3, 4 > || =√

(−3)2 + (4)2 (def of norm)

25 (by Inspection)

≈ 5 (by Calc)

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~vn =1

||~v||< − 3, 4> (given)

||~v|| ,4

||~v||

≈⟨

(approximate)

≈ 〈−0.6, 0.8〉 (by inspection)

(d) NORMALIZE them Compute and draw the corresponding normalized vector:

~v =< −3,−4 >

||~v|| = || < −3,−4 > || =√

(−3)2 + (−4)2 (def of norm)

25 (by Inspection)

≈ 5 (by Calc)

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~vn =1

||~v||< − 3,−4> (given)

||~v|| ,−4

||~v||

≈⟨

5,−4

(approximate)

≈ 〈−0.6,−0.8〉 (by inspection)

(e) NORMALIZE them Compute and draw the corresponding normalized vector:

~v =< −3,−1 >

||~v|| = || < −3,−1 > || =√

(−3)2 + (−1)2 (def of norm)

10 (by Inspection)

≈ 3.16 (by Calc)

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~vn =1

||~v||< − 3,−1> (given)

||~v|| ,−1

||~v||

≈⟨

3.16,−1

(approximate)

≈ 〈−0.95,−0.32〉 (by inspection)

(f) ~v =< a, b > (not both zero..)

Solution:

~vn =1

||~v||<a, b> (given)

||~v|| ,b

||~v||

a√a2 + b2

a2 + b2

(def of norm)

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1. Vector Dot Product

(a) Compute the following DOT product:

< 1, 2 > · < 3, 5 >

Solution:

< 1, 2 > · < 3, 5 > = (1)(3) + (2)(5) (def of the DOT)

= 3 + 10 (by Calc)

= 13 (by Calc)

(b) Compute the following DOT product:

< 3, 5 > · < 3, 5 >

Solution:

< 3, 5 > · < 3, 5 > = (3)(3) + (5)(5) (def of the DOT)

= 9 + 25 (by Calc)

= 34 (by Calc)

(c) Compute the following DOT product:

< −3, 2 > · < 1, 2 >

Solution:

< −3, 2 > · < 1, 2 > = (−3)(1) + (2)(2) (def of the DOT)

= −3 + 4 (by Calc)

= 1 (by Calc)

(d) Compute the following DOT product:

< 1, 4 > · < −3,−5 >

Solution:

< 1, 4 > · < −3,−5 > = (1)(−3) + (4)(−5) (def of the DOT)

= −3 + −20 (by Calc)

= −23 (by Calc)

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(e) Compute the following DOT product:

< 2, 3 > · < 3,−3 >

Solution:

< 2, 3 > · < 3,−3 > = (2)(3) + (3)(−3) (def of the DOT)

= 6 + −9 (by Calc)

= −3 (by Calc)

(f) Compute the following DOT product:

< 1,−4 > · < 3, 7 >

Solution:

< 1,−4 > · < 3, 7 > = (1)(3) + (−4)(7) (def of the DOT)

= 3 + −28 (by Calc)

= −25 (by Calc)

(g) < 2, 1, 3 > · < 1, 2, 1 >

Solution: 2 + 2 + 3 = 7

2. Vector Dot Product

(a) Compute the following DOT product:

(1i + 2j) · (0i + 5j)

Solution:

(1i + 2j) · (0i + 5j) = (1 < 1, 0 > +2 < 0, 1 >) · (0 < 1, 0 > +5 < 0, 1 >) (def of i and j)

=< 1, 2 > · < 0, 5 > (by Inspection)

= (1)(0) + (2)(5) (def of the DOT)

= 0 + 10 (by Calc)

= 10 (by Calc)

(b) Compute the following DOT product:

(3i + 0j) · (3i + 5j)

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Solution:

(3i + 0j) · (3i + 5j) = (3 < 1, 0 > +0 < 0, 1 >) · (3 < 1, 0 > +5 < 0, 1 >) (def of i and j)

=< 3, 0 > · < 3, 5 > (by Inspection)

= (3)(3) + (0)(5) (def of the DOT)

= 9 + 0 (by Calc)

= 9 (by Calc)

(c) Compute the following DOT product:

(−3i + 2j) · (1i + 2j)

Solution:

(−3i + 2j) · (1i + 2j) = (−3 < 1, 0 > +2 < 0, 1 >) · (1 < 1, 0 > +2 < 0, 1 >) (def of i and j)

=< −3, 2 > · < 1, 2 > (by Inspection)

= (−3)(1) + (2)(2) (def of the DOT)

= −3 + 4 (by Calc)

= 1 (by Calc)

(d) Compute the following DOT product:

(1i + 4j) · (−3i + −5j)

Solution:

(1i + 4j) · (−3i + −5j) = (1 < 1, 0 > +4 < 0, 1 >) · (−3 < 1, 0 > + − 5 < 0, 1 >) (def of i and j)

=< 1, 4 > · < −3,−5 > (by Inspection)

= (1)(−3) + (4)(−5) (def of the DOT)

= −3 + −20 (by Calc)

= −23 (by Calc)

3. Vector Dot Product to find magnitude

(a) Use the dot product to find the magnitude of the following vector:

~v =< 1, 2 >

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Solution:

||~v||2 = ~v · ~v (famous DOT product property)

=< 1, 2 > · < 1, 2 > (given)

= (1)(1) + (2)(2) (def of the DOT)

= 1 + 4 (by Calc)

....thus.... ||~v||2 = 5 (by Calc)

....then.. (assuming norm is positive).... ||~v|| =√

||~v|| ≈ 2.24

(b) Use the dot product to find the magnitude of the following vector:

~v =< 3, 5 >

Solution:

=< 3, 5 > · < 3, 5 > (given)

= (3)(3) + (5)(5) (def of the DOT)

= 9 + 25 (by Calc)

....thus.... ||~v||2 = 34 (by Calc)

||~v|| ≈ 5.83

(c) Use the dot product to find the magnitude of the following vector:

~v =< −3, 2 >

Solution:

=< −3, 2 > · < −3, 2 > (given)

= (−3)(−3) + (2)(2) (def of the DOT)

= 9 + 4 (by Calc)

....thus.... ||~v||2 = 13 (by Calc)

||~v|| ≈ 3.61

(d) Use the dot product to find the magnitude of the following vector:

~v =< 1, 4 >

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Solution:

=< 1, 4 > · < 1, 4 > (given)

= (1)(1) + (4)(4) (def of the DOT)

= 1 + 16 (by Calc)

....thus.... ||~v||2 = 17 (by Calc)

||~v|| ≈ 4.12

(e) Use the dot product to find the magnitude of the following vector:

~v =< 2, 3 >

Solution:

=< 2, 3 > · < 2, 3 > (given)

= (2)(2) + (3)(3) (def of the DOT)

= 4 + 9 (by Calc)

....thus.... ||~v||2 = 13 (by Calc)

||~v|| ≈ 3.61

(f) Use the dot product to find the magnitude of the following vector:

~v =< 1,−4 >

Solution:

=< 1,−4 > · < 1,−4 > (given)

= (1)(1) + (−4)(−4) (def of the DOT)

= 1 + 16 (by Calc)

....thus.... ||~v||2 = 17 (by Calc)

||~v|| ≈ 4.12

(g) Use the dot product to find the magnitude of the following vector:

~v =< 3,−4 >

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Solution:

=< 3,−4 > · < 3,−4 > (given)

= (3)(3) + (−4)(−4) (def of the DOT)

= 9 + 16 (by Calc)

....thus.... ||~v||2 = 25 (by Calc)

||~v|| ≈ 5

4. Vector Dot Product to find ’distance’

The DOT product provides a way to define the ’distance’ between two vectors, ~v and ~w. So long as we can definesubtraction of the vectors we can define the distance between them as follows:

dist(~v, ~w) = ||~v − ~w|| =√

(~v − ~w) · (~v − ~w)

(a) Use the dot product to find the ’distance’ between the indicated vectors.

~w =< 1, 2 > ~v =< 3, 5 >

Solution: First note that ~v − ~w =< 2, 3 >

||~v − ~w||2 = (~v − ~w) · (~v − ~w)(famous DOT product property)

=< 2, 3 > · < 2, 3 > (from given)

= (2)(2) + (3)(3) (def of the DOT)

= 4 + 9 (by Calc)

....thus.... ||~v − ~w||2 = 13 (by Calc)

....then.. (assuming the distance is positive).... ||~v − ~w|| =√

||~v − ~w|| ≈ 3.61

(b) Use the dot product to find the ’distance’ between the indicated vectors.

~w =< −3, 2 > ~v =< 1, 2 >

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=< 4, 0 > · < 4, 0 > (from given)

= (4)(4) + (0)(0) (def of the DOT)

= 16 + 0 (by Calc)

....thus.... ||~v − ~w||2 = 16 (by Calc)

||~v − ~w|| ≈ 4

(c) Use the dot product to find the ’distance’ between the indicated vectors.

~w =< 1, 4 > ~v =< −3,−5 >

Solution: First note that ~v − ~w =< −4,−9 >

=< −4,−9 > · < −4,−9 > (from given)

= (−4)(−4) + (−9)(−9) (def of the DOT)

= 16 + 81 (by Calc)

....thus.... ||~v − ~w||2 = 97 (by Calc)

||~v − ~w|| ≈ 9.85

(d) Use the dot product to find the ’distance’ between the indicated vectors.

~w =< 2, 3 > ~v =< 3,−3 >

Solution: First note that ~v − ~w =< 1,−6 >

=< 1,−6 > · < 1,−6 > (from given)

= (1)(1) + (−6)(−6) (def of the DOT)

= 1 + 36 (by Calc)

....thus.... ||~v − ~w||2 = 37 (by Calc)

||~v − ~w|| ≈ 6.08

(e) Use the dot product to find the ’distance’ between the indicated vectors.

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~w =< 1,−4 > ~v =< 3, 7 >

=< 2, 11 > · < 2, 11 > (from given)

= (2)(2) + (11)(11) (def of the DOT)

= 4 + 121 (by Calc)

....thus.... ||~v − ~w||2 = 125 (by Calc)

||~v − ~w|| ≈ 11.18

(f) Use the dot product to find the ’distance’ between the indicated vectors.

~w =< 3,−4 > ~v =< 3, 7 >

=< 0, 11 > · < 0, 11 > (from given)

= (0)(0) + (11)(11) (def of the DOT)

= 0 + 121 (by Calc)

....thus.... ||~v − ~w||2 = 121 (by Calc)

||~v − ~w|| ≈ 11

(g) Use the dot product to find the ’distance’ between the indicated vectors.

~w =< 1,−3, 5, 0, 6 > ~v =< 5, 2,−10, 3, 1 >

Solution: don’t be afraid, don’t google it, don’t ask anyone.. just you and the problem.. its on..., if itknocks you down, just get up... don’t let it beat you..

5. Vector Dot Product to find ’angle’ between two vectors

(a) Use the dot product to find the ’angle’ between the indicated vectors.

~w =< 1, 2 > ~v =< 3,−5 >

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Solution: Let us assume the angle between the vectors is between 0 and 180◦ Suppose we call such angle’θ’, then....

cos θ =~v · ~w

||~v||||~w|| (famous DOT product property)

=< 1, 2 > · < 3,−5 >

|| < 1, 2 > |||| < 3,−5 > || (given)

≈ −7

(2.236)(5.831)≈ −7

13.038(by Calc)

....thus.... cos θ = −0.54 (by Calc)

....then.. (assume the sought angle is 0 ≤ θ ≤ 180◦).... θ ≈ cos−1(−0.54)

θ ≈ 2.14 radians

≈ 122.6◦

(b) Use the dot product to find the ’angle’ between the indicated vectors.

~w =< −3, 2 > ~v =< 3, 5 >

Solution: Let us assume the angle between the vectors is between 0 and 180◦ Suppose we call such angle

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’θ’, then....

cos θ =~v · ~w

=< −3, 2 > · < 3, 5 >

|| < −3, 2 > |||| < 3, 5 > || (given)

(3.606)(5.831)≈ 1

21.027(by Calc)

....thus.... cos θ = 0.05 (by Calc)

....then.. (assume the sought angle is 0 ≤ θ ≤ 180◦).... θ ≈ cos−1(0.05)

θ ≈ 1.52 radians

≈ 87.1◦

(c) Use the dot product to find the ’angle’ between the indicated vectors.

~w =< 1, 4 > ~v =< −3,−5 >

cos θ =~v · ~w

=< 1, 4 > · < −3,−5 >

|| < 1, 4 > |||| < −3,−5 > || (given)

≈ −23

(4.123)(5.831)≈ −23

24.041(by Calc)

....thus.... cos θ = −0.96 (by Calc)

θ ≈ 2.86 radians

≈ 163.9◦

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(d) Use the dot product to find the ’angle’ between the indicated vectors.

~w =< 2, 3 > ~v =< 3,−3 >

cos θ =~v · ~w

=< 2, 3 > · < 3,−3 >

|| < 2, 3 > |||| < 3,−3 > || (given)

≈ −3

(3.606)(4.243)≈ −3

15.3(by Calc)

....thus.... cos θ = −0.2 (by Calc)

θ ≈ 1.77 radians

≈ 101.4◦

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(e) Use the dot product to find the ’angle’ between the indicated vectors.

~w =< 3, 7 > ~v =< 1,−4 >

cos θ =~v · ~w

=< 3, 7 > · < 1,−4 >

|| < 3, 7 > |||| < 1,−4 > || (given)

≈ −25

(7.616)(4.123)≈ −25

31.401(by Calc)

....thus.... cos θ = −0.8 (by Calc)

θ ≈ 2.5 radians

≈ 143.2◦

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(f) Use the dot product to find the ’angle’ between the indicated vectors.

~w =< 3, 9 > ~v =< 7, 1 >

cos θ =~v · ~w

=< 3, 9 > · < 7, 1 >

|| < 3, 9 > |||| < 7, 1 > || (given)

≈ 30

(9.487)(7.071)≈ 30

67.083(by Calc)

....thus.... cos θ = 0.45 (by Calc)

....then.. (assume the sought angle is 0 ≤ θ ≤ 180◦).... θ ≈ cos−1(0.45)

θ ≈ 1.1 radians

≈ 63◦

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(g) Use the dot product to find the ’angle’ between the indicated vectors.

~w =< −5, 2 > ~v =< 2, 5 >

cos θ =~v · ~w

=< −5, 2 > · < 2, 5 >

|| < −5, 2 > |||| < 2, 5 > || (given)

(5.385)(5.385)≈ 0

28.998(by Calc)

....thus.... cos θ = 0 (by Calc)

....then.. (assume the sought angle is 0 ≤ θ ≤ 180◦).... θ ≈ cos−1(0)

θ ≈ 1.57 radians

≈ 90◦

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(h) Use the dot product to find the ’angle’ between the indicated vectors.

~w =< 1, 7 > ~v =< 7,−1 >

cos θ =~v · ~w

=< 1, 7 > · < 7,−1 >

|| < 1, 7 > |||| < 7,−1 > || (given)

(7.071)(7.071)≈ 0

49.999(by Calc)

θ ≈ 1.57 radians

≈ 90◦

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(i) Use the dot product to find the ’angle’ between the indicated vectors.

~w =< 2, 3 > ~v =< 3,−2 >

cos θ =~v · ~w

=< 2, 3 > · < 3,−2 >

|| < 2, 3 > |||| < 3,−2 > || (given)

(3.606)(3.606)≈ 0

13.003(by Calc)

θ ≈ 1.57 radians

≈ 90◦

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(j) Use the dot product to find the ’angle’ between the indicated vectors.

~w =< 1,−3, 5, 0, 6 > ~v =< 5, 2,−10, 3, 1 >

6. Perpendicular Test by the DOT Find the Dot product for each pair of vectors, then determine if they areperpendicular.

(a) Use the dot product to find test if the indicated vectors are ’perpendicular’.

~w =< 1, 2 > ~v =< 3,−5 >

Solution: The perpendicular test is quite simple and elegant. Vector ~v is perpendicular to ~w if and only if~v · ~w = 0, or more concisely,

~v⊥~w ⇐⇒ ~v · ~w = 0

Thus we check..

~v · ~w =< 3,−5 > · < 1, 2 > (given)

= −7 (by inspection)

....therefore....~v and ~w are not perpendicular.

(b) Use the dot product to find test if the indicated vectors are ’perpendicular’.

~w =< 3, 2 > ~v =< 2,−3 >

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~v⊥~w ⇐⇒ ~v · ~w = 0

Thus we check..

~v · ~w =< 2,−3 > · < 3, 2 > (given)

= 0 (by inspection)

....therefore....~v and ~w are perpendicular.

(c) Use the dot product to find test if the indicated vectors are ’perpendicular’.

~w =< 1, 4 > ~v =< −3,−5 >

~v⊥~w ⇐⇒ ~v · ~w = 0

Thus we check..

~v · ~w =< −3,−5 > · < 1, 4 > (given)

(d) Use the dot product to find test if the indicated vectors are ’perpendicular’.

~w =< 2, 3 > ~v =< 3,−3 >

~v⊥~w ⇐⇒ ~v · ~w = 0

Thus we check..

~v · ~w =< 3,−3 > · < 2, 3 > (given)

(e) Use the dot product to find test if the indicated vectors are ’perpendicular’.

~w =< 3, 6 > ~v =< 2,−1 >

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~v⊥~w ⇐⇒ ~v · ~w = 0

Thus we check..

~v · ~w =< 2,−1 > · < 3, 6 > (given)

= 0 (by inspection)

(f) Use the dot product to find test if the indicated vectors are ’perpendicular’.

~w =< 3, 12 > ~v =< −4, 1 >

~v⊥~w ⇐⇒ ~v · ~w = 0

Thus we check..

~v · ~w =< −4, 1 > · < 3, 12 > (given)

= 0 (by inspection)

(g) Use the dot product to find test if the indicated vectors are ’perpendicular’.

~w =< −5, 2 > ~v =< 3, 5 >

~v⊥~w ⇐⇒ ~v · ~w = 0

Thus we check..

~v · ~w =< 3, 5 > · < −5, 2 > (given)

(h) Use the dot product to find test if the indicated vectors are ’perpendicular’.

~w =< 1, 7 > ~v =< 7,−2 >

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~v⊥~w ⇐⇒ ~v · ~w = 0

Thus we check..

~v · ~w =< 7,−2 > · < 1, 7 > (given)

(i) Use the dot product to find test if the indicated vectors are ’perpendicular’.

~w =< 2, 3 > ~v =< 3,−2 >

~v⊥~w ⇐⇒ ~v · ~w = 0

Thus we check..

~v · ~w =< 3,−2 > · < 2, 3 > (given)

= 0 (by inspection)

(j) test to see if perpendicular...

~w =< 1,−3, 5,−2, 6 > ~v =< 5, 0,−1, 3, 1 >

7. Projections by the DOT

(a) Use the dot product to find the ’projection’ of ~v onto ~w then draw such projection.

~v =< 1, 2 > ~w =< 5, 0 >

Solution:

proj~w~v =~v · ~w

||~w||2 ~w (famous DOT product property)

=< 1, 2 > · < 5, 0 >

|| < 5, 0 > ||2 < 5, 0 > (given)

25< 5, 0 > (by Calc)

≈ 0.2< 5, 0 > (by Calc)

≈ < 1, 0 > (by Calc)

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(b) Use the dot product to find the ’projection’ of ~v onto ~w then draw such projection.

~v =< 3, 2 > ~w =< 5, 0 >

Solution:

proj~w~v =~v · ~w

=< 3, 2 > · < 5, 0 >

|| < 5, 0 > ||2 < 5, 0 > (given)

≈ 15

25< 5, 0 > (by Calc)

≈ 0.6< 5, 0 > (by Calc)

≈ < 3, 0 > (by Calc)

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(c) Use the dot product to find the ’projection’ of ~v onto ~w then draw such projection.

~v =< 4, 2 > ~w =< 5, 0 >

Solution:

proj~w~v =~v · ~w

=< 4, 2 > · < 5, 0 >

|| < 5, 0 > ||2 < 5, 0 > (given)

≈ 20

25< 5, 0 > (by Calc)

≈ 0.8< 5, 0 > (by Calc)

≈ < 4, 0 > (by Calc)

(d) Use the dot product to find the ’projection’ of ~v onto ~w then draw such projection.

~v =< 7, 2 > ~w =< 5, 0 >

Solution:

proj~w~v =~v · ~w

=< 7, 2 > · < 5, 0 >

|| < 5, 0 > ||2 < 5, 0 > (given)

≈ 35

25< 5, 0 > (by Calc)

≈ 1.4< 5, 0 > (by Calc)

≈ < 7, 0 > (by Calc)

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(e) Use the dot product to find the ’projection’ of ~v onto ~w then draw such projection.

~v =< −3, 2 > ~w =< 5, 0 >

Solution:

proj~w~v =~v · ~w

=< −3, 2 > · < 5, 0 >

|| < 5, 0 > ||2 < 5, 0 > (given)

≈ −15

25< 5, 0 > (by Calc)

≈ −0.6< 5, 0 > (by Calc)

≈ < −3, 0 > (by Calc)

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(f) Use the dot product to find the ’projection’ of ~v onto ~w then draw such projection.

~v =< 1, 2 > ~w =< 0, 6 >

Solution:

proj~w~v =~v · ~w

=< 1, 2 > · < 0, 6 >

|| < 0, 6 > ||2 < 0, 6 > (given)

≈ 12

36< 0, 6 > (by Calc)

≈ 0.333< 0, 6 > (by Calc)

≈ < 0, 2 > (by Calc)

(g) Use the dot product to find the ’projection’ of ~v onto ~w then draw such projection.

~v =< 3, 2 > ~w =< 0, 7 >

Solution:

proj~w~v =~v · ~w

=< 3, 2 > · < 0, 7 >

|| < 0, 7 > ||2 < 0, 7 > (given)

≈ 14

49< 0, 7 > (by Calc)

≈ 0.286< 0, 7 > (by Calc)

≈ < 0, 2 > (by Calc)

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(h) Use the dot product to find the ’projection’ of ~v onto ~w then draw such projection.

~v =< 4, 2 > ~w =< 0, 2 >

Solution:

proj~w~v =~v · ~w

=< 4, 2 > · < 0, 2 >

|| < 0, 2 > ||2 < 0, 2 > (given)

4< 0, 2 > (by Calc)

≈ 1< 0, 2 > (by Calc)

≈ < 0, 2 > (by Calc)

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(i) Use the dot product to find the ’projection’ of ~v onto ~w then draw such projection.

~v =< 7, 2 > ~w =< 0, 4 >

Solution:

proj~w~v =~v · ~w

=< 7, 2 > · < 0, 4 >

|| < 0, 4 > ||2 < 0, 4 > (given)

16< 0, 4 > (by Calc)

≈ 0.5< 0, 4 > (by Calc)

≈ < 0, 2 > (by Calc)

(j) Use the dot product to find the ’projection’ of ~v onto ~w then draw such projection.

~v =< −3, 2 > ~w =< 0, 3 >

Solution:

proj~w~v =~v · ~w

=< −3, 2 > · < 0, 3 >

|| < 0, 3 > ||2 < 0, 3 > (given)

9< 0, 3 > (by Calc)

≈ 0.667< 0, 3 > (by Calc)

≈ < 0, 2 > (by Calc)

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(k) Use the dot product to find the ’projection’ of ~v onto ~w then draw such projection.

~v =< 1, 2 > ~w =< 1, 6 >

Solution:

proj~w~v =~v · ~w

=< 1, 2 > · < 1, 6 >

|| < 1, 6 > ||2 < 1, 6 > (given)

≈ 13

37< 1, 6 > (by Calc)

≈ 0.351< 1, 6 > (by Calc)

≈ < 0.35, 2.11 > (by Calc)

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(l) Use the dot product to find the ’projection’ of ~v onto ~w then draw such projection.

~v =< 3, 2 > ~w =< 7, 1 >

Solution:

proj~w~v =~v · ~w

=< 3, 2 > · < 7, 1 >

|| < 7, 1 > ||2 < 7, 1 > (given)

≈ 23

50< 7, 1 > (by Calc)

≈ 0.46< 7, 1 > (by Calc)

≈ < 3.22, 0.46 > (by Calc)

(m) Use the dot product to find the ’projection’ of ~v onto ~w then draw such projection.

~v =< 4, 2 > ~w =< 4, 2 >

Solution:

proj~w~v =~v · ~w

=< 4, 2 > · < 4, 2 >

|| < 4, 2 > ||2 < 4, 2 > (given)

≈ 20

20< 4, 2 > (by Calc)

≈ 1< 4, 2 > (by Calc)

≈ < 4, 2 > (by Calc)

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(n) Use the dot product to find the ’projection’ of ~v onto ~w then draw such projection.

~v =< 7, 2 > ~w =< 8, 1 >

Solution:

proj~w~v =~v · ~w

=< 7, 2 > · < 8, 1 >

|| < 8, 1 > ||2 < 8, 1 > (given)

≈ 58

65< 8, 1 > (by Calc)

≈ 0.892< 8, 1 > (by Calc)

≈ < 7.14, 0.89 > (by Calc)

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(o) Use the dot product to find the ’projection’ of ~v onto ~w then draw such projection.

~v =< 3, 2 > ~w =< 9, 3 >

Solution:

proj~w~v =~v · ~w

=< 3, 2 > · < 9, 3 >

|| < 9, 3 > ||2 < 9, 3 > (given)

≈ 33

90< 9, 3 > (by Calc)

≈ 0.367< 9, 3 > (by Calc)

≈ < 3.3, 1.1 > (by Calc)

(p) find projection~v =< 1,−3, 5,−2, 6 > onto~w =< 5, 0,−1, 3, 1 >

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Neuroimmunological and Neuroenergetic Aspects in Exercise …eir-isei.de/2019/eir-2019-008-article.pdf · 2019-02-18 · (ROS), both exercise-induced muscle damage (47,48) and the

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