If a- and b are two non-zero vectors- - a - b -2 is equal to which of the following-A- a2 b2-a-b-2B- a b-a-b-2C- a2 b2-a-b-2D- a2 b2-a-b-

The correct option is A a^2 b^2 (a.b)^2Let|a| = a, |b| = b and θ be the angle between them. |a×b|^2 = (ab sinθ )^2 = a^2 b^2 sin ^2 θ = a^2 b^2(1 cos^2 θ) = ...

You visited us 4 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Multiplication with Vectors

If a⃗ and b a...

Question

If $\to a$ and $\to b$ are two non-zero vectors, $| \to a \times \to b |^{2}$ is equal to which of the following?

a^{2} b^{2} - (\to a . \to b)^{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

a^{2} b^{2} + (\to a . \to b)^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

a b - (\to a . \to b)^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

a^{2} b^{2} - (\to a . \to b)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $a^{2} b^{2} - (\to a . \to b)^{2}$
Let $| \to a | = a, | \to b | = b$ and $θ$ be the angle between them.
$| \to a \times \to b |^{2} = (a b sin θ)^{2} = a^{2} b^{2} {sin}^{2} θ$
$= a^{2} b^{2} (1 - c o s^{2} θ) = a^{2} b^{2} - (a . b cos θ)^{2}$
$= a^{2} b^{2} - (\to a . \to b)^{2}$

Suggest Corrections

Similar questions

Q. Find the value of

| \to a \times \to b |^{2} .

Q. For any two vectors a and b, P.T

| a \times b |^{2} = (a . a) (b .) - (a . b)^{2} = a^{2} b^{2} - (a . b)^{2}

Q. Prove that

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

Q. Given

x = 1 + a + a^{2} + . . . . \infty

and

y = 1 + b + b^{2} . . . . \infty

where

a

and

b

are proper fraction

1 + a b + a^{2} b^{2} + . . . . \infty

equals to ..........

The length of the chord intercepted by the circle $x^{2} + y^{2} = r^{2}$ on the line $\frac{x}{a} + \frac{y}{b} = 1$

Join BYJU'S Learning Program

If →a and →b are two non-zero vectors, |→a×→b|2 is equal to which of the following?

The correct option is A a2b2−(→a.→b)2Let|→a|=a,|→b|=b and θ be the angle between them. |→a×→b|2=(absinθ)2=a2b2sin2θ =a2b2(1−cos2θ)=a2b2−(a.bcosθ)2 =a2b2−(→a.→b)2

If $\to a$ and $\to b$ are two non-zero vectors, $| \to a \times \to b |^{2}$ is equal to which of the following?