Test: Vector Algebra: Dot and Cross Product(22 Nov) - JEE MCQ

As we know that,

Area of triangle = ½(a * b)
a = (1, 0, -2)   b = (2, 3, 1)
= i(0 + 6) + j(-4 - 1) + k(3 - 0)
= 6i - 5j + 3k
|a * b| = (36 + 25 + 9)^½
|a * b| = (70)^½
Area of triangle = ½(a * b)
= [(70)^½]/2

 i.(j * k) + j.(k * i) + k(i * j)
= i.(i) + j.(j) + k.(k)
= 1 + 1 + 1 
= 3

A = {(1, -3, 4)}    B = {(3, 1, -2)}
A * B = {(i^(6-4) - j^(-2 -12) + k^ (1+9)} 
= 2i^ + 14j^ + 10k^
Area = [(2)² + (14)² + (10)²]^½
= (300)^½
= 10(3)^½ sq unit

a = 6i - 3j + 2k      b = 2i + j - 2k
a.b = 12 - 3 - 4 = 5
|a| = [(6)² + (-3)² + (2)²]^1/2 
|a| = [36 + 9 + 4]^½
|a| = (49)^½
|a| = 7
|b| = [(2)² + (1)² + (-2)²]^½
|b| = [4 + 1 + 4]^½
|b| = 3
Cosθ = (a.b)/|a||b|
= 5/(7)(3)
= 5/21
θ = cos^-1(5/21)

 |a - b|² = |a|² + |b|² - 2|a||b|
|a - b|²  = (3)² + (2)² - 2(5)
|a - b|²  = 9 + 4 - 10
|a - b|²  = 3 
|a - b|   = (3)^½.

Projection = (A.B)/|B|
= [(i + 2j + k) . (2i + 3j + 2k)]/[(2)² + (3)² + (2)²]^½
= (2 + 6 + 2)/[4 + 9 + 4]^½
= 10/(17)^1/2

A sequence is a function whose domain is the set of natural numbers or a subset of the natural numbers. We usually use the symbol an to represent a sequence, where n is a natural number and an is the value of the function on n. A sequence may be finite or infinite.