The distance of the plane vec{r} = (hat{i} - | vedclass

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(A) The equation of the plane is given in parametric form as $\vec{r} = \vec{a} + \lambda\vec{b} + \mu\vec{c}$,where $\vec{a} = \hat{i} - \hat{j}$,$\v

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$\frac{7}{\sqrt{38}}$ units

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(A) The equation of the plane is given in parametric form as $\vec{r} = \vec{a} + \lambda\vec{b} + \mu\vec{c}$,where $\vec{a} = \hat{i} - \hat{j}$,$\vec{b} = \hat{i} + \hat{j} + \hat{k}$,and $\vec{c} = \hat{i} - 2\hat{j} + 3\hat{k}$.To find the normal vector $\vec{n}$ to the plane,we calculate the cross product $\vec{n} = \vec{b} 	imes \vec{c}$:$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ 1 & -2 & 3 \end{vmatrix} = \hat{i}(3 - (-2)) - \hat{j}(3 - 1) + \hat{k}(-2 - 1) = 5\hat{i} - 2\hat{j} - 3\hat{k}$.The equation of the plane in scalar form is $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$,which is $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$.Calculating $\vec{a} \cdot \vec{n} = (\hat{i} - \hat{j}) \cdot (5\hat{i} - 2\hat{j} - 3\hat{k}) = (1)(5) + (-1)(-2) + (0)(-3) = 5 + 2 = 7$.So,the equation of the plane is $5x - 2y - 3z = 7$.The distance $d$ of the plane $Ax + By + Cz = D$ from the origin $(0, 0, 0)$ is given by $d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$.Here,$A = 5, B = -2, C = -3$,and $D = 7$.$d = \frac{|7|}{\sqrt{5^2 + (-2)^2 + (-3)^2}} = \frac{7}{\sqrt{25 + 4 + 9}} = \frac{7}{\sqrt{38}}$ units.

The distance of the plane $\vec{r} = (\hat{i} - \hat{j}) + \lambda(\hat{i} + \hat{j} + \hat{k}) + \mu(\hat{i} - 2\hat{j} + 3\hat{k})$ from the origin is

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