For vectors a- b- Prove that -a-b-2 - -a-2-b-2 - -a- - b-2

We know that, |a⃗ ×b⃗|^2 =|a⃗|^2 |b⃗|^2 sin^2θ |a⃗ ×b⃗|^2 =|a⃗|^2 |b⃗|^2 (1 cos^2θ) |a⃗ ×b⃗|^2 =|a⃗|^2 |b⃗|^2 |a⃗|^2 |b⃗|^2cos^2θ #160; ...

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Applications of Dot Product

For vectors ...

Question

For vectors $\to a$ & $\to b$ . Prove that
$| \to a \times \to b |^{2} = | \to a |^{2} | \to b |^{2} - | \to a . \to b |^{2}$

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Solution

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Similar questions

\to a, \to b

{∣ ∣ \to a \times \to b ∣ ∣}^{2} = ∣ ∣ ∣ \begin{matrix} \to a . \to a & \to a . \to b \to b . \to a & \to b . \to b \end{matrix} ∣ ∣ ∣

\to a

\to b

\frac{{∣ ∣ \to a + \to b ∣ ∣}^{2} - {∣ ∣ \to a - \to b ∣ ∣}^{2}}{{∣ ∣ \to a + \to b ∣ ∣}^{2} + {∣ ∣ \to a - \to b ∣ ∣}^{2}}

[\to a + \to b, \to b + \to c, \to c + \to a] = 2 [\to a, \to b, \to c]

| \to a + \to b | = | \to a - \to b |

(\to a, \to b) = \frac{π}{2}

\to a, \to b, \to a + \to b

(\to a, \to b) = \frac{2 π}{3}

\to a

\to b

\to c

\to a + \to b + \to c = 0

\to a \times \to b = \to b \times \to c = \to c \times \to a

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