If bar{r}=(2-lambda+mu) hat{i}+(1-mu) hat{j}+ | vedclass

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(A) Given the vector equation of the plane: $\bar{r}=(2-\lambda+\mu) \hat{i}+(1-\mu) \hat{j}+(2-3 \lambda+2 \mu) \hat{k}$.We can rewrite this as: $\ba

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$3 x+y-z=5$

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(A) Given the vector equation of the plane: $\bar{r}=(2-\lambda+\mu) \hat{i}+(1-\mu) \hat{j}+(2-3 \lambda+2 \mu) \hat{k}$.We can rewrite this as: $\bar{r}=(2 \hat{i}+\hat{j}+2 \hat{k})+\lambda(-\hat{i}-3 \hat{k})+\mu(\hat{i}-\hat{j}+2 \hat{k})$.This represents a plane passing through the point $(2, 1, 2)$ and parallel to the vectors $\bar{a} = -\hat{i}-3 \hat{k}$ and $\bar{b} = \hat{i}-\hat{j}+2 \hat{k}$.The normal vector $\bar{n}$ to the plane is given by the cross product $\bar{a} 	imes \bar{b}$:$\bar{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1 & 0 & -3 \ 1 & -1 & 2 \end{vmatrix} = \hat{i}(0-3) - \hat{j}(-2+3) + \hat{k}(1-0) = -3 \hat{i} - \hat{j} + \hat{k}$.Alternatively, using the vectors $\bar{a} = \hat{i}+3 \hat{k}$ and $\bar{b} = \hat{i}-\hat{j}+2 \hat{k}$, the normal is $\bar{n} = (\hat{i}+3 \hat{k}) 	imes (\hat{i}-\hat{j}+2 \hat{k}) = 3 \hat{i} + \hat{j} - \hat{k}$.The equation of the plane is $(\bar{r} - \bar{r}_0) \cdot \bar{n} = 0$, where $\bar{r}_0 = 2 \hat{i}+\hat{j}+2 \hat{k}$.$(x \hat{i} + y \hat{j} + z \hat{k} - (2 \hat{i}+\hat{j}+2 \hat{k})) \cdot (3 \hat{i} + \hat{j} - \hat{k}) = 0$.$3(x-2) + 1(y-1) - 1(z-2) = 0$.$3x - 6 + y - 1 - z + 2 = 0$.$3x + y - z = 5$.

If $\bar{r}=(2-\lambda+\mu) \hat{i}+(1-\mu) \hat{j}+(2-3 \lambda+2 \mu) \hat{k}$ is the vector equation of a plane, then the equivalent cartesian equation of the plane is

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