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Question
Show that the plane whose vector equation is \[\vec{r} \cdot \left( \hat{i} + 2 \hat{j} - \hat{k} \right) = 3\] contains the line whose vector equation is \[\vec{r} = \hat{i} + \hat{j} + \lambda\left( 2 \hat{i} + \hat{j} + 4 \hat{k} \right) .\]
Solution
\[\text{ The line } \vec{r} = \left( \hat{i} + \hat{j} + 0 \hat{k} \right) + \lambda \left( 2 \hat{i} + \hat{j} + 4 \hat{k} \right) . . . . (1) \text{ passes through a point whose position vector is } \vec{a} = \hat{i} + \hat{j} + 0 \hat{k} \text{ and is parallel to the vector} \vec{b} =2 \hat{i} + \hat{j} + 4 \hat{k} . \]
\[\text{ If the plane } \vec{r} .\left( \hat{i} + 2 \hat{j} - \hat{k} \right)=3 \text{ contains the given line, then } \]
\[(1) \text{ it should passes through the point } \hat{i} + \hat{j} + 0 \hat{k} \]
\[(2) \text{ it should be parallel to the line} \]
\[\text{ Now } ,\left( \hat{i} + \hat{j} + 0 \hat{k} \right).\left( \hat{i} + 2 \hat{j} - \hat{k} \right)= 1 + 2 = 3\]
\[\text{ So, the plane passes through the point } \hat{i} + \hat{j}+ 0 \hat{k} . \]
\[\text{ The normal vector to the given plane is } \vec{n} = \hat{i} + 2 \hat{j} - \hat{k} \]
\[\text{ We observe that} \]
\[ \vec{b} . \vec{n} = \left( 2 \hat{i} + \hat{j} + 4 \hat{k} \right) . \left( \hat{i} + 2 \hat{j} - \hat{k} \right) = 2 + 2 - 4 = 0\]
\[\text{ Therefore, the plane is parallel to the line.} \]
\[\text{ Hence, the given plane contains the given line } .\]
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