The value of (overrightarrow A + overrightarr | vedclass

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(D) We use the distributive property of the cross product:$(\vec A + \vec B) 	imes (\vec A - \vec B) = \vec A 	imes \vec A - \vec A 	imes \vec B +

Vedclass · Accepted Answer

$2(\overrightarrow B 	imes \overrightarrow A)$

Vedclass · Answer

(D) We use the distributive property of the cross product:$(\vec A + \vec B) 	imes (\vec A - \vec B) = \vec A 	imes \vec A - \vec A 	imes \vec B + \vec B 	imes \vec A - \vec B 	imes \vec B$Since the cross product of any vector with itself is zero ($\vec A 	imes \vec A = 0$ and $\vec B 	imes \vec B = 0$):$= 0 - (\vec A 	imes \vec B) + (\vec B 	imes \vec A) - 0$Using the anticommutative property of the cross product $(\vec A 	imes \vec B = -(\vec B 	imes \vec A))$:$= -(\vec A 	imes \vec B) + (\vec B 	imes \vec A) = (\vec B 	imes \vec A) + (\vec B 	imes \vec A) = 2(\vec B 	imes \vec A)$

The value of $(\overrightarrow A + \overrightarrow B ) \times (\overrightarrow A - \overrightarrow B )$ is

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