The vector equation of the plane r = (2 hat{i | vedclass

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(A) The given equation of the plane is $r = (2 \hat{i} + \hat{k}) + \lambda(\hat{i}) + \mu(\hat{i} + 2 \hat{j} - 3 \hat{k})$.This plane passes through

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$2$

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(A) The given equation of the plane is $r = (2 \hat{i} + \hat{k}) + \lambda(\hat{i}) + \mu(\hat{i} + 2 \hat{j} - 3 \hat{k})$.This plane passes through the point $a = 2 \hat{i} + \hat{k}$ and is parallel to the vectors $b = \hat{i}$ and $c = \hat{i} + 2 \hat{j} - 3 \hat{k}$.The scalar product form of the plane is $r \cdot (b 	imes c) = a \cdot (b 	imes c)$.First,calculate the normal vector $n = b 	imes c$:$n = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 0 & 0 \ 1 & 2 & -3 \end{vmatrix} = \hat{i}(0 - 0) - \hat{j}(-3 - 0) + \hat{k}(2 - 0) = 3 \hat{j} + 2 \hat{k}$.Wait,re-calculating the cross product: $\hat{i}(0) - \hat{j}(-3) + \hat{k}(2) = 3 \hat{j} + 2 \hat{k}$.Actually,the question specifies $r \cdot (3 \hat{i} + 2 \hat{k}) = \alpha$. Let us re-check the cross product: $b = \hat{i} = (1, 0, 0)$,$c = (1, 2, -3)$.$b 	imes c = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 0 & 0 \ 1 & 2 & -3 \end{vmatrix} = \hat{i}(0) - \hat{j}(-3) + \hat{k}(2) = 3 \hat{j} + 2 \hat{k}$.There seems to be a typo in the provided question's normal vector. Assuming the normal is $3 \hat{j} + 2 \hat{k}$,then $\alpha = a \cdot n = (2 \hat{i} + \hat{k}) \cdot (3 \hat{j} + 2 \hat{k}) = (2)(0) + (0)(3) + (1)(2) = 2$.Thus,$\alpha = 2$.

The vector equation of the plane $r = (2 \hat{i} + \hat{k}) + \lambda(\hat{i}) + \mu(\hat{i} + 2 \hat{j} - 3 \hat{k})$ in scalar product form is $r \cdot (3 \hat{i} + 2 \hat{k}) = \alpha$,then $\alpha = \dots$

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