A plane pi_1 passing through the point 3 hat{ | vedclass

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(C) The vector equation of a plane passing through point $\vec{a}$ with normal vector $\vec{n}$ is $(\vec{r}-\vec{a}) \cdot \vec{n} = 0$,which simplif

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$3$

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(C) The vector equation of a plane passing through point $\vec{a}$ with normal vector $\vec{n}$ is $(\vec{r}-\vec{a}) \cdot \vec{n} = 0$,which simplifies to $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$.For plane $\pi_1$,$\vec{a}_1 = 3 \hat{i}-7 \hat{j}+5 \hat{k}$ and $\vec{n}_1 = \hat{i}+2 \hat{j}-2 \hat{k}$.$\vec{r} \cdot (\hat{i}+2 \hat{j}-2 \hat{k}) = (3 \hat{i}-7 \hat{j}+5 \hat{k}) \cdot (\hat{i}+2 \hat{j}-2 \hat{k}) = 3 - 14 - 10 = -21$.The normal form is $\vec{r} \cdot \hat{n} = p$. Here,$|\vec{n}_1| = \sqrt{1^2+2^2+(-2)^2} = 3$.Dividing by $3$,$\vec{r} \cdot \frac{\hat{i}+2 \hat{j}-2 \hat{k}}{3} = \frac{-21}{3} = -7$. Thus,$p_1 = |-7| = 7$.For plane $\pi_2$,$\vec{a}_2 = 2 \hat{i}+7 \hat{j}-8 \hat{k}$ and $\vec{n}_2 = 3 \hat{i}+2 \hat{j}+6 \hat{k}$.$\vec{r} \cdot (3 \hat{i}+2 \hat{j}+6 \hat{k}) = (2 \hat{i}+7 \hat{j}-8 \hat{k}) \cdot (3 \hat{i}+2 \hat{j}+6 \hat{k}) = 6 + 14 - 48 = -28$.Here,$|\vec{n}_2| = \sqrt{3^2+2^2+6^2} = \sqrt{9+4+36} = 7$.Dividing by $7$,$\vec{r} \cdot \frac{3 \hat{i}+2 \hat{j}+6 \hat{k}}{7} = \frac{-28}{7} = -4$. Thus,$p_2 = |-4| = 4$.Therefore,$p_1 - p_2 = 7 - 4 = 3$.

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