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Question

(B) The line of intersection of the planes $r \cdot (3i - j + k) = 1$ and $r \cdot (i + 4j - 2k) = 2$ is parallel to the vector $n = n_1 	imes n_2$,w

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$r \cdot (2i - 7j - 13k) = 1$

Vedclass · Answer

(B) The line of intersection of the planes $r \cdot (3i - j + k) = 1$ and $r \cdot (i + 4j - 2k) = 2$ is parallel to the vector $n = n_1 	imes n_2$,where $n_1 = 3i - j + k$ and $n_2 = i + 4j - 2k$.Calculating the cross product: $n = n_1 	imes n_2 = \begin{vmatrix} i & j & k \ 3 & -1 & 1 \ 1 & 4 & -2 \end{vmatrix} = i(2 - 4) - j(-6 - 1) + k(12 + 1) = -2i + 7j + 13k$.The plane is perpendicular to this line,so the normal vector to the plane is $n = -2i + 7j + 13k$.The plane passes through the point $a = i + 2j - k$.The equation of the plane is $(r - a) \cdot n = 0$,which is $r \cdot n = a \cdot n$.$r \cdot (-2i + 7j + 13k) = (i + 2j - k) \cdot (-2i + 7j + 13k) = -2 + 14 - 13 = -1$.Multiplying by $-1$,we get $r \cdot (2i - 7j - 13k) = 1$.

The vector equation of the plane passing through the point $i + 2j - k$ and perpendicular to the line of intersection of the planes $r \cdot (3i - j + k) = 1$ and $r \cdot (i + 4j - 2k) = 2$ is

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