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.

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  • C. \(6x-3y+z=0\)

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 \(0\).

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Review the concept: Lines and planes in space

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.

\[\begin{aligned}\mathbf r(t)&=\mathbf r_0+t\,\mathbf d\\\mathbf n\cdot(\mathbf r-\mathbf r_0)&=0\end{aligned}\]

More practice problems: lines and planes in space

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