a and b are non-zero non-collinear vectors such that |a| = 2, a.b = 1 and the angle between a and b is \pi/3. If r is any vectors satisfying r.a = 2, r.b = 8, (r + 2a — 10b) . (a x b) = 6 and r + 2a-10a = λ (a x b), then λ =

NIMCET 2019 Question Paper with solutions

Question : 22 of 120

Marks: +1, -0

a and b are non-zero non-collinear vectors such that |a| = 2, a.b = 1 and the angle between a and b is

π/3

. If r is any vectors satisfying r.a = 2, r.b = 8, (r + 2a — 10b) . (a x b) = 6 and r + 2a-10a = λ (a x b), then λ =

Solution:

Given,

|a|=2

and and

a⋅b=1

Angle between a and b is

π/3

∴\;a⋅b={|a||b|}\cosπ/3

1=2×{|b|}×1/2

⇒|b|=1

(r+2a-10b)⋅(a×b)=6

and

(r+2a-10b)=λ(a×b)

∴\;λ(a×b)⋅(a×b)=6

λ|a×b|^2=6

λ|a^2||b^2|\sin^2π/3=6

λ×4×1×3/4=6

λ=2

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