The Cartesian equation of the plane r = (1 + | vedclass

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(C) Given the vector equation of the plane: $r = (1 + \lambda - \mu )i + (2 - \lambda )j + (3 - 2\lambda + 2\mu )k$This can be rewritten as: $r = (i +

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$2x + z = 5$

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(C) Given the vector equation of the plane: $r = (1 + \lambda - \mu )i + (2 - \lambda )j + (3 - 2\lambda + 2\mu )k$This can be rewritten as: $r = (i + 2j + 3k) + \lambda (i - j - 2k) + \mu (-i + 2k)$This represents a plane passing through the point $a = i + 2j + 3k$ and parallel to the vectors $b = i - j - 2k$ and $c = -i + 2k$.The normal vector $n$ to the plane is given by the cross product $n = b 	imes c$:$n = \begin{vmatrix} i & j & k \ 1 & -1 & -2 \ -1 & 0 & 2 \end{vmatrix} = i(-2 - 0) - j(2 - 2) + k(0 - 1) = -2i - k$The vector equation of the plane is $(r - a) \cdot n = 0$, which implies $r \cdot n = a \cdot n$.Calculating $a \cdot n$: $(i + 2j + 3k) \cdot (-2i - k) = (1)(-2) + (2)(0) + (3)(-1) = -2 - 3 = -5$.Thus, $r \cdot (-2i - k) = -5$, or $r \cdot (2i + k) = 5$.Substituting $r = xi + yj + zk$, we get $(xi + yj + zk) \cdot (2i + k) = 5$, which simplifies to $2x + z = 5$.

The Cartesian equation of the plane $r = (1 + \lambda - \mu )i + (2 - \lambda )j + (3 - 2\lambda + 2\mu )k$ is

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