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JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Vector Algebra

Let

a

and

b

be two vectors such that

∣ a + b ∣^{2} = ∣ a ∣^{2} + 2∣ b ∣^{2}

,

a \cdot b = 3

and

∣ a \times b ∣^{2} = 75

. Then

∣ a ∣^{2}

is equal to ____.

Answer: 14

Solution

From the first condition:

∣ a + b ∣^{2} = ∣ a ∣^{2} + 2∣ b ∣^{2}

Expanding the left side:

∣ a ∣^{2} + ∣ b ∣^{2} + 2 a \cdot b = ∣ a ∣^{2} + 2∣ b ∣^{2}

Simplifying:

∣ b ∣^{2} = 2 a \cdot b

Given

a \cdot b = 3

, we get:

∣ b ∣^{2} = 6

Using the cross product identity:

∣ a \times b ∣^{2} = ∣ a ∣^{2} ∣ b ∣^{2} - (a \cdot b)^{2}

Substituting known values:

75 = ∣ a ∣^{2} (6) - (3)^{2}

75 = 6∣ a ∣^{2} - 9

6∣ a ∣^{2} = 84

∣ a ∣^{2} = 14

JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Integer answer

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