Question

Let ⃗a = î + jˆ + √(2) kˆ, ⃗b = b₁î + b₂jˆ + √(2) kˆ and ⃗c = 5î + jˆ + √(2) kˆ
be three vectors such that the projection vector of ⃗b
on ⃗a
is ⃗a . If ⃗a + ⃗b
is perpendicular to ⃗c, then |⃗b|
is equal to:
(A) √22 (B) 4
(C) √(32) (D) 6

Answer 1

|b| is equal to the:

(D) 6

This can be calculated as follows:

• We know that, the projection of b on a is = (b.a) / |a|

• Thus, a = [ ( b₁iˆ + b₂jˆ + √2kˆ ) ( iˆ + jˆ + √2kˆ ) ] / 1²+1²+(√2)²

                     = ( b₁ + b₂ + 2 ) / √4

                     = ( b₁ + b₂ + 2 ) / 2

• Also, the projection of b on a is = |a|

• Thus, ( b₁ + b₂ + 2 ) / 2 =  √1²+1²+(√2)²   = 2                                  

• Therefore, b₁ + b₂ = 2

• So,  a+b = ( iˆ + jˆ + √2kˆ ) + ( b₁iˆ + b₂jˆ + √2kˆ )

                      = ( b₁ + 1 ) iˆ + ( b₂ + 1 ) jˆ + 2√2kˆ

• Since, ( a+b ) is perpendicular to c, therefore,  (a+b) .c =0

• Thus, {( b₁ + 1 ) iˆ + ( b₂ + 1 ) jˆ + 2√2kˆ} ( 5iˆ + jˆ + √2kˆ ) = 0

                                                       5( bb₁ + 1 ( b₂ + 1 ) + 2√2) = 0

                                                                                   5b₁ + b₂ = −10

• Therefore, b₁ = −3 and b₂ = 5

        b = −3iˆ + 5jˆ + √2kˆ

• Hence, |b| = √(−3)²+(5)²+(√2)²   = √36                          

                                                            = 6