VECTORS
Introduction of Vectors
Vector: A vector is a directed line segment.
Vector in Plane:
Consider a point P(x, y) in xy-plane.
Position vector is a vector which describes the position of P(x, y) w.r.t. origin.
Vector OP is given as:
OP⃗ = xî + yĵ
where î = Unit vector along x-axis ⇒ |î| = 1
ĵ = Unit vector along y-axis ⇒ |ĵ| = 1
Vector in Space:
For a point P(x, y, z) in space:
OP⃗ = xî + yĵ + zḵ
where ḵ = Unit vector along z-axis
Magnitude / Norm / Length of a Vector
The distance of the point P from origin is called Magnitude of a vector.
If OP⃗ = xî + yĵ + zḵ
then |OP⃗| = √(x² + y² + z²)
Note: We can represent a vector by using bold letters u, v, w etc.
xî + yĵ ⇒ u = [x, y] or u = [x, y, z] ⇒ xî + yĵ + zḵ
Example: Find the length of the vector.
u = [1, 2, 3]
|u| = √((1)² + (2)² + (3)²) = √(1+4+9) = √14
Example: Find the distance between the vectors:
u = [1, 2, 3]; v = [-4, 3, 5]
Sol: Find u − v
|u − v| = √((−4−1)² + (3−2)² + (5−3)²)
= √(25 + 1 + 4)
= √30 ANS
Unit Vector
, A vector of magnitude one is called Unit Vector.
Unit Vector = Vector / Magnitude of Vector
Note:
(i) î = [1,0,0]; ĵ = [0,1,0]; ḵ = [0,0,1]
(ii) î, ĵ and ḵ are mutually perpendicular to each other.
Exercise 5.1 — Attempt Questions 1 to 6
Q#7: Find all values of c if |u| = 3
Given: u = [2, c, 0]
Sol: Given |u| = 3
⇒ √((2)² + (c)² + (0)²) = 3
Squaring both sides:
4 + c² = 9
c² = 5
⇒ c = ±√5 ANS
Q#2: Find cosθ = ?
u = [1, 2]; v = [-4, -5]
Sol:
u.v = (1)(−4) + (2)(−5) = −4 − 10 = −14
|u| = √((1)² + (2)²) = √5
|v| = √((−4)² + (−5)²) = √(16+25) = √41
cosθ = u.v / |u||v| = −14 / (√5 · √41) ANS
Dot Product or Inner Product of Two Vectors
(Vector)(Vector) = Scalar
Definition:
For two non-zero vectors u and v:
u.v = |u||v| cosθ
where θ is the angle between u and v.
Facts:
(1) cosθ = u.v / |u||v|
(2) u.v = v.u
(3) u.u = u²
Introduction of Vectors
Vector: A vector is a directed line segment.
Vector in Plane:
Consider a point P(x, y) in xy-plane.
Position vector is a vector which describes the position of P(x, y) w.r.t. origin.
Vector OP is given as:
OP⃗ = xî + yĵ
where î = Unit vector along x-axis ⇒ |î| = 1
ĵ = Unit vector along y-axis ⇒ |ĵ| = 1
Vector in Space:
For a point P(x, y, z) in space:
OP⃗ = xî + yĵ + zḵ
where ḵ = Unit vector along z-axis
Magnitude / Norm / Length of a Vector
The distance of the point P from origin is called Magnitude of a vector.
If OP⃗ = xî + yĵ + zḵ
then |OP⃗| = √(x² + y² + z²)
Note: We can represent a vector by using bold letters u, v, w etc.
xî + yĵ ⇒ u = [x, y] or u = [x, y, z] ⇒ xî + yĵ + zḵ
Example: Find the length of the vector.
u = [1, 2, 3]
|u| = √((1)² + (2)² + (3)²) = √(1+4+9) = √14
Example: Find the distance between the vectors:
u = [1, 2, 3]; v = [-4, 3, 5]
Sol: Find u − v
|u − v| = √((−4−1)² + (3−2)² + (5−3)²)
= √(25 + 1 + 4)
= √30 ANS
Unit Vector
, A vector of magnitude one is called Unit Vector.
Unit Vector = Vector / Magnitude of Vector
Note:
(i) î = [1,0,0]; ĵ = [0,1,0]; ḵ = [0,0,1]
(ii) î, ĵ and ḵ are mutually perpendicular to each other.
Exercise 5.1 — Attempt Questions 1 to 6
Q#7: Find all values of c if |u| = 3
Given: u = [2, c, 0]
Sol: Given |u| = 3
⇒ √((2)² + (c)² + (0)²) = 3
Squaring both sides:
4 + c² = 9
c² = 5
⇒ c = ±√5 ANS
Q#2: Find cosθ = ?
u = [1, 2]; v = [-4, -5]
Sol:
u.v = (1)(−4) + (2)(−5) = −4 − 10 = −14
|u| = √((1)² + (2)²) = √5
|v| = √((−4)² + (−5)²) = √(16+25) = √41
cosθ = u.v / |u||v| = −14 / (√5 · √41) ANS
Dot Product or Inner Product of Two Vectors
(Vector)(Vector) = Scalar
Definition:
For two non-zero vectors u and v:
u.v = |u||v| cosθ
where θ is the angle between u and v.
Facts:
(1) cosθ = u.v / |u||v|
(2) u.v = v.u
(3) u.u = u²
