If bar{a} = bar{i} - 2bar{j} - 2bar{k} and ba | vedclass

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(D) Given $\bar{a} = \bar{i} - 2\bar{j} - 2\bar{k}$ and $\bar{b} = 2\bar{i} + \bar{j} + 2\bar{k}$.We need to compute $(\bar{a} + 2\bar{b}) 	imes (3\b

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$14\bar{i} + 42\bar{j} - 35\bar{k}$

Vedclass · Answer

(D) Given $\bar{a} = \bar{i} - 2\bar{j} - 2\bar{k}$ and $\bar{b} = 2\bar{i} + \bar{j} + 2\bar{k}$.We need to compute $(\bar{a} + 2\bar{b}) 	imes (3\bar{a} - \bar{b})$.Using the distributive property of the cross product:$(\bar{a} + 2\bar{b}) 	imes (3\bar{a} - \bar{b}) = \bar{a} 	imes (3\bar{a}) - \bar{a} 	imes \bar{b} + (2\bar{b}) 	imes (3\bar{a}) - (2\bar{b}) 	imes \bar{b}$.Since $\bar{v} 	imes \bar{v} = 0$, we have $\bar{a} 	imes \bar{a} = 0$ and $\bar{b} 	imes \bar{b} = 0$.So, the expression simplifies to: $0 - (\bar{a} 	imes \bar{b}) + 6(\bar{b} 	imes \bar{a}) - 0$.Since $\bar{b} 	imes \bar{a} = -(\bar{a} 	imes \bar{b})$, we get:$-(\bar{a} 	imes \bar{b}) - 6(\bar{a} 	imes \bar{b}) = -7(\bar{a} 	imes \bar{b})$.Now, calculate $\bar{a} 	imes \bar{b} = \begin{vmatrix} \bar{i} & \bar{j} & \bar{k} \ 1 & -2 & -2 \ 2 & 1 & 2 \end{vmatrix} = \bar{i}(-4 - (-2)) - \bar{j}(2 - (-4)) + \bar{k}(1 - (-4)) = -2\bar{i} - 6\bar{j} + 5\bar{k}$.Finally, $-7(\bar{a} 	imes \bar{b}) = -7(-2\bar{i} - 6\bar{j} + 5\bar{k}) = 14\bar{i} + 42\bar{j} - 35\bar{k}$.

If $\bar{a} = \bar{i} - 2\bar{j} - 2\bar{k}$ and $\bar{b} = 2\bar{i} + \bar{j} + 2\bar{k}$ are two vectors, then $(\bar{a} + 2\bar{b}) \times (3\bar{a} - \bar{b}) = $

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