A force F applied on a body is written as F = | vedclass

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(D) Given the expression $F = (\hat{n} \cdot F) \hat{n} + G$,we can isolate $G$ as $G = F - (\hat{n} \cdot F) \hat{n}$.Using the vector triple product

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$(\hat{n} 	imes F) 	imes \hat{n}$

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(D) Given the expression $F = (\hat{n} \cdot F) \hat{n} + G$,we can isolate $G$ as $G = F - (\hat{n} \cdot F) \hat{n}$.Using the vector triple product identity $A 	imes (B 	imes C) = B(A \cdot C) - C(A \cdot B)$,we evaluate $(\hat{n} 	imes F) 	imes \hat{n}$.Note that $(\hat{n} 	imes F) 	imes \hat{n} = -\hat{n} 	imes (\hat{n} 	imes F)$.Applying the identity: $-\hat{n} 	imes (\hat{n} 	imes F) = -[\hat{n}(\hat{n} \cdot F) - F(\hat{n} \cdot \hat{n})]$.Since $\hat{n}$ is a unit vector,$\hat{n} \cdot \hat{n} = 1$.Thus,$-\hat{n}(\hat{n} \cdot F) + F(1) = F - (\hat{n} \cdot F) \hat{n}$.Comparing this with our expression for $G$,we find $G = (\hat{n} 	imes F) 	imes \hat{n}$.

$A$ force $F$ applied on a body is written as $F = (\hat{n} \cdot F) \hat{n} + G$,where $\hat{n}$ is a unit vector. The vector $G$ is equal to

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