Let overline{A}=2 hat{i}+hat{k},overline{B}=h | vedclass

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(C) Given $\overline{R} 	imes \overline{B} = \overline{C} 	imes \overline{B}$,we can write $(\overline{R} - \overline{C}) 	imes \overline{B} = 0$.

Vedclass · Accepted Answer

$-\hat{i}-8 \hat{j}+2 \hat{k}$

Vedclass · Answer

(C) Given $\overline{R} 	imes \overline{B} = \overline{C} 	imes \overline{B}$,we can write $(\overline{R} - \overline{C}) 	imes \overline{B} = 0$. This implies that $(\overline{R} - \overline{C})$ is parallel to $\overline{B}$,so $\overline{R} - \overline{C} = k\overline{B}$ for some scalar $k$. Thus,$\overline{R} = \overline{C} + k\overline{B}$. Given $\overline{R} \cdot \overline{A} = 0$,we substitute $\overline{R}$: $(\overline{C} + k\overline{B}) \cdot \overline{A} = 0 \Rightarrow \overline{C} \cdot \overline{A} + k(\overline{B} \cdot \overline{A}) = 0$. Calculate the dot products: $\overline{A} \cdot \overline{C} = (2 \hat{i} + \hat{k}) \cdot (4 \hat{i} - 3 \hat{j} + 7 \hat{k}) = 2(4) + 0(-3) + 1(7) = 8 + 7 = 15$. $\overline{A} \cdot \overline{B} = (2 \hat{i} + \hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k}) = 2(1) + 0(1) + 1(1) = 2 + 1 = 3$. Substituting these values: $15 + k(3) = 0 \Rightarrow 3k = -15 \Rightarrow k = -5$. Now,find $\overline{R}$: $\overline{R} = \overline{C} - 5\overline{B} = (4 \hat{i} - 3 \hat{j} + 7 \hat{k}) - 5(\hat{i} + \hat{j} + \hat{k}) = (4-5)\hat{i} + (-3-5)\hat{j} + (7-5)\hat{k} = -\hat{i} - 8\hat{j} + 2\hat{k}$.

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