Vector Algebra

Answer :

Work done = w = F ∙ AB

                             AB = B – A = (2 – 2)i + (1 + 1)j + 0k = 2j

                             ∴ w = (2i + j – k) ∙ (2j) = 0 + 2– 0 = 2

Answer :

  |a – b|² + |b – c|² + |c – a|²      = 2(|a|² + |b|² + |c|²) – (ab + bc + ca)

                                                          = 2[(1 + 1 + 1) – (ab + bc + ca)]

                                                          = 2[3 – (ab + bc + ca)]                      (1)

                   but    0 ≤ |a + b + c|²       = |a|² + |b|² + |c|² + 2(ab + bc + ca)

                                                = 3 + 2(ab + bc + ca)

                             ∴ ab + bc + ca ≥ [(– 3) / 2]                                                      (2)

                   putting (2) in (1),

                             |a – b|² + |b – c|² + |c – a|² ≤ 6 + 3 = 9

                   The bound 9 is attained if a + b + c = 0.

Answer :

a ∙ b < 0

                             |a ∙ b| = |a × b|

                             ∴ ab sinθ = ab cosθ (– 1) (as a ∙ b < 0)

                             ∴ tanθ = – 1

                             ∴ θ = [(– π) / 4] i.e. (3π / 4)

Answer :

To find: x × (y ∙ z)

                             x, y, z ∈ R³

                   it is obvious that x × (y ∙ z) is not possible.

Answer :

i × (x × i) + j × (x × j) + k × (x × k)

                             = (i ∙ i)x – (i ∙ x)i + (j ∙ j)x – (j ∙ x)j + (k ∙ k)x – (k ∙ x)k

                             = x – (i ∙ x)i + x – (j ∙ x)j + x – (k ∙ x) ∙ k

                             = 3x – [(i ∙ x)i + (j ∙ x)j + (k ∙ x) ∙ k]

                   We know (i ∙ x) ∙ i + (j ∙ x) ∙ j + (k ∙ x) ∙ k = x

                   ∴ expression becomes 3x – x = 2x.