The Cartesian equation of the plane bar{r}=(h | vedclass

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(D) The given equation of the plane is in the form $\bar{r} = \bar{a} + \lambda \bar{A} + \mu \bar{B}$,where $\bar{a} = \hat{i} - \hat{j}$,$\bar{A} =

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$5 x-2 y-3 z-7=0$

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(D) The given equation of the plane is in the form $\bar{r} = \bar{a} + \lambda \bar{A} + \mu \bar{B}$,where $\bar{a} = \hat{i} - \hat{j}$,$\bar{A} = \hat{i} + \hat{j} + \hat{k}$,and $\bar{B} = \hat{i} - 2\hat{j} + 3\hat{k}$.The normal vector $\bar{n}$ to the plane is given by the cross product of the two vectors $\bar{A}$ and $\bar{B}$:$\bar{n} = \bar{A} 	imes \bar{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ 1 & -2 & 3 \end{vmatrix}$$= \hat{i}(3 - (-2)) - \hat{j}(3 - 1) + \hat{k}(-2 - 1)$$= 5\hat{i} - 2\hat{j} - 3\hat{k}$.The Cartesian equation of the plane is given by $(\bar{r} - \bar{a}) \cdot \bar{n} = 0$,which is $\bar{r} \cdot \bar{n} = \bar{a} \cdot \bar{n}$.Calculating $\bar{a} \cdot \bar{n} = (\hat{i} - \hat{j}) \cdot (5\hat{i} - 2\hat{j} - 3\hat{k}) = (1)(5) + (-1)(-2) + (0)(-3) = 5 + 2 = 7$.Thus,the Cartesian equation is $5x - 2y - 3z = 7$,or $5x - 2y - 3z - 7 = 0$.

The Cartesian equation of the plane $\bar{r}=(\hat{i}-\hat{j})+\lambda(\hat{i}+\hat{j}+\hat{k})+\mu(\hat{i}-2 \hat{j}+3 \hat{k})$ is

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