Exercise · ZW-LINES-AND-PLANES-IN-SPACE-004 · ★★☆ Intermediate

What is the equation of the plane through \((1,0,0)\), \((0,1,0)\) and \((0,0,1)\)?

Choose one option.

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  • B. \(x+y+z=1\)

Explanation

Two vectors in the plane are \(\langle-1,1,0\rangle\) and \(\langle-1,0,1\rangle\); their cross product is \(\langle1,1,1\rangle\), the normal. With the point \((1,0,0)\) we get \(x+y+z=1\).

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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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