- a-b -2- a-b 2 is equal to

The correct option is C | a |^2 | b |^2 | a×b|^2+ ( a⃗.b⃗ )^2 = |a|^2 |b|^2 sin^2θ + |a|^2 |b|^2 cos^2θ = |a|^2 |b|^2 (sin^2θ + cos^2θ) ...

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Multiplication with Vectors

| a×b |2+ a.b...

Question

${∣ ∣ ∣ \to a \times \to b ∣ ∣ ∣}^{2} + {(\to a . \to b)}^{2}$ is equal to

0

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

{∣ ∣ \to a ∣ ∣}^{2} {∣ ∣ ∣ \to b ∣ ∣ ∣}^{2}

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

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

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

1

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

Open in App

Solution

Suggest Corrections

Similar questions

∣ ∣ ¯ ¯ ¯ a ∣ ∣ = ∣ ∣ ¯ ¯ b ∣ ∣ = 2

{(¯ ¯ ¯ a \times ¯ ¯ b)}^{2} + {(¯ ¯ ¯ a . ¯ ¯ b)}^{2} =

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

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

p =

| \to a + \to b |^{2} + | \to a - \to b |^{2}

| \to a |^{2}

| \to b |^{2}

a + b + c = 0

\frac{1}{b^{2} + c^{2} - a^{2}} + \frac{1}{c^{2} + a^{2} - b^{2}} + \frac{1}{a^{2} + b^{2} - c^{2}}

Join BYJU'S Learning Program