If overrightarrow{a}=2 hat{i}-5 hat{j}+8 hat{ | vedclass

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(B) Given vectors are $\overrightarrow{a}=2 \hat{i}-5 \hat{j}+8 \hat{k}$ and $\overrightarrow{b}=7 \hat{i}-5 \hat{j}+3 \hat{k}$.First,calculate $2 \ov

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$1400$

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(B) Given vectors are $\overrightarrow{a}=2 \hat{i}-5 \hat{j}+8 \hat{k}$ and $\overrightarrow{b}=7 \hat{i}-5 \hat{j}+3 \hat{k}$.First,calculate $2 \overrightarrow{a}-3 \overrightarrow{b}$:$2 \overrightarrow{a}-3 \overrightarrow{b} = 2(2 \hat{i}-5 \hat{j}+8 \hat{k}) - 3(7 \hat{i}-5 \hat{j}+3 \hat{k}) = (4-21) \hat{i} + (-10+15) \hat{j} + (16-9) \hat{k} = -17 \hat{i} + 5 \hat{j} + 7 \hat{k}$.Next,calculate $4 \overrightarrow{a}+\overrightarrow{b}$:$4 \overrightarrow{a}+\overrightarrow{b} = 4(2 \hat{i}-5 \hat{j}+8 \hat{k}) + (7 \hat{i}-5 \hat{j}+3 \hat{k}) = (8+7) \hat{i} + (-20-5) \hat{j} + (32+3) \hat{k} = 15 \hat{i} - 25 \hat{j} + 35 \hat{k}$.Now,compute the cross product $(2 \overrightarrow{a}-3 \overrightarrow{b}) 	imes(4 \overrightarrow{a}+\overrightarrow{b})$:$(2 \overrightarrow{a}-3 \overrightarrow{b}) 	imes(4 \overrightarrow{a}+\overrightarrow{b}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -17 & 5 & 7 \ 15 & -25 & 35 \end{vmatrix}$$= \hat{i}(5 	imes 35 - 7 	imes (-25)) - \hat{j}((-17) 	imes 35 - 7 	imes 15) + \hat{k}((-17) 	imes (-25) - 5 	imes 15)$$= \hat{i}(175 + 175) - \hat{j}(-595 - 105) + \hat{k}(425 - 75)$$= 350 \hat{i} + 700 \hat{j} + 350 \hat{k}$.Comparing this with $x \hat{i}+y \hat{j}+z \hat{k}$,we get $x=350, y=700, z=350$.Therefore,$x+y+z = 350+700+350 = 1400$.

If $\overrightarrow{a}=2 \hat{i}-5 \hat{j}+8 \hat{k}$ and $\overrightarrow{b}=7 \hat{i}-5 \hat{j}+3 \hat{k}$ are two vectors and $(2 \overrightarrow{a}-3 \overrightarrow{b}) \times(4 \overrightarrow{a}+\overrightarrow{b})=x \hat{i}+y \hat{j}+z \hat{k}$,then $x+y+z=$

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