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JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Vector Algebra

Let

a

and

b

be two vectors such that

∣2 a + 3 b ∣ = ∣3 a + b ∣

and the angle between

a

and

b

is

6 0^{\circ}

. If

\frac{1}{8} a

is a unit vector, then

∣ b ∣

is equal to:

(A)6
(B)5✓ Correct
(C)8
(D)4

Answer: (B)

Solution

Squaring

∣2 a + 3 b ∣ = ∣3 a + b ∣

gives

4∣ a ∣^{2} + 12 a \cdot b + 9∣ b ∣^{2} = 9∣ a ∣^{2} + 6 a \cdot b + ∣ b ∣^{2}

, so

6 a \cdot b + 8∣ b ∣^{2} = 5∣ a ∣^{2}

. With

a \cdot b = ∣ a ∣∣ b ∣ cos 6 0^{\circ} = \frac{1}{2} ∣ a ∣∣ b ∣

and

∣ a ∣ = 8

, this is

24∣ b ∣ + 8∣ b ∣^{2} = 320

, i.e.

∣ b ∣^{2} + 3∣ b ∣ - 40 = 0

, so

(∣ b ∣ + 8) (∣ b ∣ - 5) = 0

. Since a magnitude is non-negative,

∣ b ∣ = 5

.

JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Single correct option

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