The Cartesian equation of the plane whose vec | vedclass

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(D) The given vector equation is $\vec{r} = (1+\lambda-\mu) \hat{i} + (2-\lambda) \hat{j} + (3-2 \lambda+2 \mu) \hat{k}$.Rearranging the terms,we get:

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$2x+z=5$

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(D) The given vector equation is $\vec{r} = (1+\lambda-\mu) \hat{i} + (2-\lambda) \hat{j} + (3-2 \lambda+2 \mu) \hat{k}$.Rearranging the terms,we get:$\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - \hat{j} - 2\hat{k}) + \mu(-\hat{i} + 2\hat{k})$.This is in the form $\vec{r} = \vec{a} + \lambda\vec{b} + \mu\vec{c}$,where $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$,$\vec{b} = \hat{i} - \hat{j} - 2\hat{k}$,and $\vec{c} = -\hat{i} + 2\hat{k}$.The normal vector $\vec{n}$ to the plane is given by $\vec{n} = \vec{b} 	imes \vec{c}$:$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & -2 \ -1 & 0 & 2 \end{vmatrix} = \hat{i}(-2 - 0) - \hat{j}(2 - 2) + \hat{k}(0 - 1) = -2\hat{i} - \hat{k}$.The equation of the plane is $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$,which implies $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$.$\vec{a} \cdot \vec{n} = (\hat{i} + 2\hat{j} + 3\hat{k}) \cdot (-2\hat{i} - \hat{k}) = (1)(-2) + (2)(0) + (3)(-1) = -2 - 3 = -5$.So,$\vec{r} \cdot (-2\hat{i} - \hat{k}) = -5$,or $\vec{r} \cdot (2\hat{i} + \hat{k}) = 5$.Substituting $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$,we get $(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (2\hat{i} + \hat{k}) = 5$,which simplifies to $2x + z = 5$.

The Cartesian equation of the plane whose vector equation is $\vec{r}=(1+\lambda-\mu) \hat{i}+(2-\lambda) \hat{j}+(3-2 \lambda+2 \mu) \hat{k}$,where $\lambda, \mu$ are scalars,is:

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