exercize

Exercise 7.3 (Solutions)

MathCity.org

Calculus and Analytic Geometry, MATHEMATICS 12

Merging man and maths

Available online @ http://www.mathcity.org, Version: 1.0.0

Question # 1
ˆ j ˆ
(i)
u = 3i + ˆ − k ,

ˆ j ˆ v = 2i − ˆ + k

u = (3)2 + (1) 2 + (−1)2 = 9 + 1 + 1 = 11 v = (2) 2 + (−1)2 + (1)2 = 4 + 1 + 1 = 6
ˆ j ˆ u ⋅ v = 3iˆ + ˆ − k ⋅ 2i − ˆ + k j ˆ

(

)(

)

= (3)(2) + (1)(−1) + (−1)(1) = 6 − 1 − 1 = 4 u ⋅ v = u v cosθ

Now

⇒ cosθ =

u⋅v u v

4
11 × 6

=

⇒ cosθ =

(ii)

Do yourself as above

(iii)

ˆ u = [ −3,5] = −3i + 5 ˆ , j Now do yourself as above

(iv)

ˆ u = [1, −3,1] = i − 3 ˆ + k j ˆ

4
66

Question # 2
ˆ ˆ
(i)
a=i −k ,

v = [ 6, − 2] = 6iˆ − 2 ˆ j ˆ
, v = 2i + 4 ˆ + k j ˆ

Now do yourself as (i)

b= ˆ+k j ˆ

a = (1) 2 + (0)2 + (−1)2 = 1 + 1 = 2 b = (0) 2 + (1)2 + (1)2 = 1 + 1 = 2
ˆ ˆ a ⋅ b = i − k ⋅ ˆ + k = (1)(0) + (0)(1) + (−1)(1) = 0 + 0 − 1 = −1 j ˆ

(

Since

)(

)

a ⋅ b = a b cosθ

a⋅b
−1
= b 2 a⋅b −1
Also projection of b along a = b cosθ =
=
a
2
So projection of a along b = a cosθ =

(ii)

Do yourself as above

Question # 3
(i)
Do yourself as (ii) below
ˆ
(ii) u = α iˆ + 2α ˆ − k , v = iˆ + α ˆ + 3k j ˆ j Since u and v are perpendicular therefore u .v = 0
ˆ
⇒ α iˆ + 2α ˆ − k ⋅ i + α ˆ + 3k = 0 j ˆ ˆ j (

)(

)

⇒ (α )(1) + (2α )(α ) + (−1)(3) = 0
⇒ α + 2α 2 − 3 = 0 ⇒ 2α 2 + α − 3 = 0
⇒ 2α 2 + 3α − 2α − 3 = 0 ⇒ α ( 2α + 3) − 1( 2α + 3) = 0

⇒ ( 2α + 3)(α − 1) = 0
⇒ 2α + 3 = 0 or α − 1 = 0
3
⇒ α =− or α =1
2

FSC-II / Ex. 7.3 - 2

Question # 4
Given vertices: A(1, −1,0) , B (−2,2,1) and C (0,2, z ) uuu r
ˆ
ˆ
CA = (1 − 0) $ + (−1 − 2) ˆ + (0 − z ) k = $ − 3 ˆ − zk i j i j uuu r
ˆ
ˆ
CB = (−2 − 0) $ + (2 − 2) ˆ + (1 − z )k = −2 $ + (1 − z )k i j i uuu r uuu r uuu uuu r r
Now CA is ⊥ to CB therefore CA ⋅ CB = 0
ˆ
ˆ
⇒ $ − 3 ˆ − zk ⋅ −2 $ + (1 − z )k = 0 i j i (