Exercise · ZW-LINES-AND-PLANES-IN-SPACE-001 · ★☆☆ Basic

What is the equation of the plane through \((1,2,3)\) with normal vector \(\langle2,-1,4\rangle\)?

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  • C. \(2x-y+4z=12\)

Explanation

The equation is \(\mathbf n\cdot(\mathbf r-\mathbf r_0)=0\), i.e. \(2(x-1)-(y-2)+4(z-3)=0\). Simplifying gives \(2x-y+4z=2-2+12=12\).

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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

ZW-LINES-AND-PLANES-IN-SPACE-002 →