Exercise · ZW-LINES-AND-PLANES-IN-SPACE-009 · ★★★ Advanced
What is the equation of the plane through the origin containing the directions \(\langle1,2,0\rangle\) and \(\langle0,1,3\rangle\)?
Choose one option.
A \(6x+3y+z=0\) B \(6x-3y+z=1\) C \(6x-3y+z=0\) D \(x+2y+3z=0\)
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Explanation The normal is the cross product : \(\langle1,2,0\rangle\times\langle0,1,3\rangle=\langle6,-3,1\rangle\). Since the plane passes through the origin, the constant term is \(0\).
A line in space is determined by a point \(\mathbf r_0\) and a direction vector \(\mathbf d\); a plane, by a point \(\mathbf r_0\) and a normal vector \(\mathbf n\) perpendicular to it.
\[\begin{aligned}\mathbf r(t)&=\mathbf r_0+t\,\mathbf d\\\mathbf n\cdot(\mathbf r-\mathbf r_0)&=0\end{aligned}\]
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