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L1: VECTORS AND SCALARS

Example (1): Represent graphically:

(a) a force of 10 lb in a direction 30° north of east

(b) a force of 15 lb in a direction 30 ° east of north.

Choosing the unit of magnitude shown, the required vectors are as indicated above.

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Example (2): An automobile travels 3 miles due north, then 5 miles northeast.

Represent these displacements graphically and determine the resultant

displacement

a. graphically.

b. analytically.

Vector OP or A represents displacement of 3 mi due north.

Vector PQ or B represents displacement of 5 mi north east.

Vector OQ or C represents the resultant displacement or

sum of vectors A and B, i.e. C = A+B. This, is the triangle law of vector

addition.

The resultant vector OQ can also be obtained by constructing the diagonal of the

parallelogram OPQR having vectors OP =A and OR (equal to vector PQ or B) as

sides. This is the parallelogram law of vector addition.

(a) Graphical Determination of Resultant. Lay off the 1 mile unit on

vector OQ to find the magnitude 7.4 mi (approximately).

Angle EOQ=61.5°, using a protractor.

Then vector OQ has magnitude 7.4 mi and direction 61.5 °

north of east.

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(b) Analytical Determination of Resultant. From triangle OPQ, denoting

the magnitudes of A, B. C by A, B, C, we have by the law of cosines

Ex. Show that addition of vectors is commutative, i.e. A + B = B + A. See Fig. (b)

below.

OP + PQ = OQ or A + B = C,

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and OR + RQ = OQ or B + A = C.

Then A + B = B + A .

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E.x. Show that the addition of vectors is associative, i.e. A + (B+C) = (A+B) + C

.

OP + PQ= OQ =(A + B), And PQ + QR= PR =(B + C).

OP + PR = OR = D, i. e. A + (B + C) = D .

OQ + QR = OR = D, i.e. (A + B) + C = D.

Then A + (B + C) = (A + B) + C.

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Figure Distributive Law for vector addition.

Figure Distributive law for scalar multiplication

Example:

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Example:

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L3: The Dot and Cross Product

THE DOT OR SCALAR PRODUCT:

PROJECTION OF VECTORS

Scalar projection

Vector Projection

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Ex. Evaluate each of the following:

(a) i.i= |𝑖||𝑖| cos 0o =(1)(1)(1) = 1

(b) i.k= |𝑖||𝑘|cos 90o =(1)(1)(0) = 0

(c) k.j= 𝑘|𝑗|cos 90o =(1)(1)(0) = 0

(d) j.(2i-3j+k) =2j.i-3j.j+j.k=0- 3 + 0 = -3

(e) (2i - j).(3i + k) = 2i.(3i + k) – j.(3i + k) = 6i.i + 2i.k - 3j.i – j.k = 6 + 0 - 0 - 0 = 6

Ex. If A = A1 i + A2 j + A3 k and B = B1 i + B2 j + B3 k, prove that A.B = A1 B1 +

A2 B2 + A3 B3

A.B = (A1 i +A2 j +A3 k) . (B1 i +B2 j +B3 k)

= A1 i. (B1 i +B2 j +B3 k)+ A2 j.(B1 i +B2 j +B3 k)+ A3 k.(B1 i +B2 j +B3 k)

=A1B1 i.i + A1B2 i.j + A1B3 i.k + A2B1 j.i + A2B2 j.j+ A2B3 j.k + A3B1 k.i + A3B2 k.j

+ A3B3 k.k

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= A1B1 + A2B2 + A3B3

since i.i = j.j = k.k = 1 and all other dot products are zero.

Ex. If A = A1 i + A2 j + A3 k, show that A = √𝐴. 𝐴 = √𝐴12 + 𝐴2

2 + 𝐴32

A.A= (A)(A) cos 0° = A2. Then A = √𝐴. 𝐴.

Also, A.A = (A1 i +A2 j +A3 k) (A1 i +A2 j +A3k)

=(A1)( A1) + (A2)(A2) + (A3)(A3) = 𝐴12 + 𝐴2

2 + 𝐴32

by previous Problem, taking B = A.

Then A= √𝐴. 𝐴 = √𝐴12 + 𝐴2

2 + 𝐴32 is the magnitude of A. Sometimes A. A is

written A2 .

Ex. Find the angle between A = 2i + 2j -k and B = 6i - 3j + 2k .

A.B = AB cos θ, A = √(22) + (22) + (−12) =3,

B =√(62) + (−32) + (22) = 7

A.B = (2)(6) + (2)(-3) + (-1)(2) = 12 - 6 - 2 = 4

Then cos θ = 𝐴.𝐵

𝐴𝐵=

4

(3)(7)=

4

21= 0.1905 and θ= 79

o approximately.

THE CROSS OR VECTOR PRODUCT

Ex. If A x B = 0 and if A and B are not zero, show that A is parallel to B.

If AxB =AB sine θ u =0, then sin θ = 0 and θ = 0° or 180°.

Ex. Show that |𝐴 × 𝐵|2 + |𝐴. 𝐵|2 = |𝐴|2|𝐵|2

= |𝐴 × 𝐵|2 + |𝐴. 𝐵|2 = |𝐴𝐵 𝑠𝑖𝑛𝜃 𝑢|2 + |𝐴𝐵 𝑐𝑜𝑠𝜃|2 = 𝐴2𝐵2𝑠𝑖𝑛2𝜃 + 𝐴2𝐵2𝑐𝑜𝑠2𝜃

=𝐴2𝐵2 = |𝐴|2|𝐵|2

Ex. Evaluate each of the following.

(a) i x j = k (f) j x j = 0

(b) j x k = i (g) i x k = -k x i = -j

(c) k x i = j (h) (2j) x (3k) = 6 j x k = 6i

(d) k x j = -j x k = - i (i) (3i) x (-2k) = - 6 i x k = 6j

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(e) i xi = 0 (j) 2j x i - 3k = -2k - 3k = -5k

Ex. If A = Al i + A2 j +A3 k and B = B1i + B2j + B3k , prove that A x B

=|𝑖 𝑗 𝑘

𝐴1 𝐴2 𝐴3

𝐵1 𝐵2 𝐵3

|

A x B = (Al i + A2 j + A3 k) x (B1 i + B2 j + B3 k)

= Ali x (B1i + B2j + B3k) + A2j x (B1i + B2j + B3k) + A3k x (B1i + B2j +B3k)

= A1B1 i x i +A1B2ixj +A1B3 i x k +A2B1 j x i +A2B2 j x j +A2B3 j x k +A3B1 k x i

+A3B2 k x j +A3B3 k x k

=(A2B3 - A3B2) i + (A3B1 - A1B3) j + (A1B2 - A2B1) k =|𝑖 𝑗 𝑘

𝐴1 𝐴2 𝐴3

𝐵1 𝐵2 𝐵3

|

Ex. If A = 2i - 3j - k and B = i + 4j - 2k, find (a) A x B, (b) B x A, (c) (A

+ B) x (A - B).

(a) A x B = (2i - 3j - k) x (i + 4j - 2k) =|𝑖 𝑗 𝑘2 −3 −11 4 −2

|

𝑖 |−3 −14 −2

| − 𝑗 |2 −11 −2

| + 𝑘 |2 −31 4

|=10 i+3 j+11 k

Another Method.

(2i - 3j -k) x (i + 4j -2k) = 2i x (i + 4j - 2k) - 3j x (i + 4j -2k) - k x (i + 4j - 2k)

= 2i x i+ 8i x j- 4i x k- 3j x i- 12j x j+ 6j x k - k x i - 4k x j + 2k x k

= 0 + 8k + 4j + 3k - 0 + 6i - j + 4i + 0 = 10i+ 3j +11k

(b) B x A = (i +4j - 2k) x (2i-3j--k) = |𝑖 𝑗 𝑘1 4 −22 −3 −1

|

𝑖 |4 −2

−3 −1| − 𝑗 |

1 −22 −1

| + 𝑘 |1 42 −3

|== -10i - 3j - 1l k.

Comparing with (a), A x B = - B x A. Note that this is equivalent to the theorem:

If two rows of a determinant are interchanged, the determinant changes sign.

(c) A+B = (2i-3j-k) + (i+4j-2k) = 3i + j - 3k

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A-B = (2i-3j-k)-(i+4j-2k) = i-7j +k

Then (A + B) x (A - B) = (3i+j-3k) x (i - 7j + k) =|𝑖 𝑗 𝑘3 1 −31 −7 1

|

𝑖 |1 −3

−7 1| − 𝑗 |

3 −31 1

| + 𝑘 |3 11 −7

|= -20i - 6j -22k.

Another Method.

(A + B) x (A - B) = A x (A- B) + B x (A - B)

= AxA- AxB +BxA - BxB = 0- AxB – AxB -0 = -2A X B

= - 2 (10i + 3j + Ilk) =- 20i - 6j - 22k, using (a).

avoid ambiguity.

TRIPLE PRODUCTS:

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L5: Vector Differentiation

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