Let a vector $\vec{a}$ has a magnitude 9. Let a vector $\vec{b}$ be such that for every $(x, y) \in R \times R-\{(0,0)\}$, the vector ( $x \vec{a}+y \vec{b}$ ) is perpendicular to the vector ( $6 y \vec{a}-18 x \vec{b}$ ). Then the value of $|\vec{a} \times \vec{b}|$ is equal to:

JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Vector Algebra

Let a vector

a

has a magnitude 9. Let a vector

b

be such that for every

(x, y) \in R \times R - {(0, 0)}

, the vector (

x a + y b

) is perpendicular to the vector (

6 y a - 18 x b

). Then the value of

∣ a \times b ∣

is equal to:

Answer: (B)

Given

∣ a ∣ = 9

and

(x a + y b) \cdot (6 y a - 18 x b) = 0

for all

(x, y) \neq = (0, 0)

. Expanding:

6 x y ∣ a ∣^{2} - 18 x^{2} (a \cdot b) + 6 y^{2} (a \cdot b) - 18 x y ∣ b ∣^{2} = 0

. Rearranging:

6 x y (∣ a ∣^{2} - 3∣ b ∣^{2}) + (a \cdot b) (6 y^{2} - 18 x^{2}) = 0

. For this to hold for all

x, y

, both coefficients must be zero:

∣ a ∣^{2} = 3∣ b ∣^{2}

and

a \cdot b = 0

. Now

∣ a \times b ∣^{2} = ∣ a ∣^{2} ∣ b ∣^{2} - (a \cdot b)^{2} = ∣ a ∣^{2} \cdot \frac{∣ a ∣ ^{2}}{3} = \frac{∣ a ∣ ^{4}}{3} = \frac{8 1 ^{2}}{3} = 2187

. Therefore

∣ a \times b ∣ = \frac{81}{3} = 273

JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Single correct option