Calculus and Vectors – How to get an A+

7.4 Dot Product of Algebraic Vectors A Dot Product for Standard Unit Vectors The dot product of the standard unit vectors is given by: r r r r r r i ⋅ i =1 j ⋅ j =1 k ⋅ k =1 r r r r r r i ⋅ j =0 j ⋅k =0 k ⋅i = 0 B Dot Product for two Algebraic Vectors The dot product of two algebraic vectors r r r r a = (a x , a y , a z ) = a x i + a y j + a z k and r r r r b = (b x , b y , b z ) = b x i + b y j + b z k is given by: r r a ⋅ b = a x bx + a y b y + a z bz Proof: r r r r r r r r a ⋅ b = (a x i + a y j + a z k ) ⋅ (bx i + b y j + bz k ) r r r r r r = (a x bx )(i ⋅ i ) + (a x b y (i ⋅ j ) + (a x bz )(i ⋅ k ) + r r r r r r + (a y bx )( j ⋅ i ) + (a y b y ( j ⋅ j ) + (a y bz )( j ⋅ k ) + r r r r r r + (a z bx )(k ⋅ i ) + (a z b y (k ⋅ j ) + (a z bz )(k ⋅ k )
= a x bx + a y b y + a z bz

Proof: r r r r i ⋅ i =|| i || || i || cos 0° = (1)(1)(1) = 1 r r r r i ⋅ j =|| i || || j || cos 90° = (1)(1)(0) = 0

r Ex 1. For each case, find the dot product of the vectors a r and b . r r a) a = (1,−2,0) , b = (0,−1,2) r r a ⋅ b = (1)(0) + (−2)(−1) + (0)(2) = 2 r r r r r r r b) a = −i + 2 j , b = i − 2 j − k r r a ⋅ b = (−1)(1) + (2)(−2) + (0)(−1) = −1 − 4 = −5 r r r r r c) a = (−1,1,−1) , b = −i + 2 j − 2k r r a ⋅ b = (−1)(−1) + (1)(2) + (−1)(−2) = 1 + 2 + 2 = 5

C Angle between two Vectors r r r The angle θ = ∠(a , b ) between two vectors a and r b (when positioned tail to tail) is given by: r r a x bx + a y b y + a z bz a ⋅b cosθ = r r = | a || b | a x 2 + a y 2 + a z 2 bx 2 + b y 2 + bz 2 Notes: r r 1. If cosθ = 1 then a ↑↑ b (vectors are parallel and have same direction). r r 2. If cosθ = −1 then a ↑↓ b (vectors are parallel but have opposite direction). r r 3. If cosθ = 0 then a ⊥ b (vectors are perpendicular to each other or orthogonal).

r Ex 2. For each case, find the angle between the vectors a r and b . r r a) a = (1,−2,−1) , b = (0,−1,2) r r a ⋅b (1)(0) + (−2)(−1) + (−1)(2) cosθ = r r = =0 2 | a || b | 1 + (−2) 2 + (−1) 2 0 2 + (−1) 2 + 2 2 r r ∴θ = cos −1 0 = 90° (a ⊥ b )

r r r r r r b) a = −i − 2k , b = −2 j + k r r a ⋅b (−1)(0) + (0)(−2) + (−2)(1) −2 = cosθ = r r = 5 | a || b | (−1) 2 + 0 2 + (−2) 2 0 2 + (−2) 2 + 12
∴θ = cos −1 (−2 / 5) = 113.58°

4. If cosθ > 0 then 0° < θ < 90° ( θ is an acute angle).

Ex 3. Find a non zero vector perpendicular to each of the r r vectors a = (1,5,−1) and b = (−3,1,2) . r r r Let v = ( x, y, z ) be a vector perpendicular to both a and b . So: r r a ⋅ v = 0 ⇒ x + 5 y − z = 0 (1) r r b ⋅ v = 0 ⇒ −3x + y + 2 z = 0 (2) (1) ⇒ z = x + 5 y (3) (2) ⇒ −3 x + y + 2( x + 5 y ) = 0 ⇒ 11 y = x (3) ⇒ z = 11y + 5 y = 16 y r v = ( x, y, z ) = (11y, y,16 y ) = y (11,1,16) r ∴ v = y (11,1,16), y ∈ R \ {0}

5. If cosθ < 0 then 90° < θ < 180° ( θ is an obtuse angle).

7.4 The Dot Product of Algebraic Vectors ©2010 Iulia & Teodoru Gugoiu - Page 1 of 2

Calculus and Vectors – How to get an A+

Ex 4. A triangle is defined by three points A(0,1,2) , B(1,0,2) , and C (−1,2,0) . Find the angles ∠A of this triangle.

Ex 5. Find the angles between the vector r r r r a = −2i + j + 3k and the coordinate axes. r r r r r a a ⋅i ∠(a , x − axis ) = ∠(a , i ) = cos −1 r r = cos −1 rx || a || || a || || i || r −2 ∴ ∠(a , x − axis ) = cos −1 ≅ 122.31° 14 r r ay r r r a⋅ j ∠(a , y − axis ) = ∠(a , j ) = cos −1 r r = cos −1 r || a || || a || || j || r 1 ∴ ∠(a , y − axis ) = cos −1 ≅ 74.5° 14 r r r r r a⋅k −1 −1 a z ∠(a , z − axis ) = ∠(a , k ) = cos r r r = cos || a || || a || || k || r 3 ∴ ∠(a , z − axis ) = cos −1 ≅ 36.7° 14

AB = (1,0,2) − (0,1,2) = (1,−1,0) AC = (−1,2,0) − (0,1,2) = (−1,1,−2) AB ⋅ AC −1 −1+ 0 −2 −1

cos A =

|| AB || || AC ||
−1

=

2 6

=

12

=

3

∴ ∠A = cos (−1 / 3 ) ≅ 125.26°

Ex 6. For what values of k are the vectors r r a = (k ,−2,3) and b = (2,2k − 6,6) a) perpendicular (orthogonal)? r r a ⋅ b = 0 ⇒ 2k − 2(2k − 6) + 18 = 0 ⇒ 2k − 4k + 12 + 18 = 0 ⇒ 30 = 2k ⇒∴ k = 15 b) parallel (collinear)? r r a = λb ⇒ (k ,−2,3) = λ (2,2k − 6,6) ⇒
⎧k = 2λ ⎪ ⎨− 2 = λ ( 2 k − 6) ⎪3 = 6λ ⇒ λ = 0.5 ⎩ k = 2(0.5) = 1 − 2 = 0.5(2 × 1 − 6) ⇒ −2 = −2 (true) ∴k = 1

c) in opposite direction? r r The vectors a and b are in opposite direction if r r there exists λ < 0 such that a = λb . But, according r r to part b) if a = λb then λ = 0.5 > 0 . r r Therefore the vectors a and b cannot be in opposite direction for any real value of the parameter k . Reading: Nelson Textbook, Pages 379-385 Homework: Nelson Textbook: Page #2c, 6d, 7a, 10, 13, 14, 18, 19

7.4 The Dot Product of Algebraic Vectors ©2010 Iulia & Teodoru Gugoiu - Page 2 of 2