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Cross Products, Lines, and Planes · Tutorial 99 of 999

Line-and-Plane Construction Problems

Combine point, direction, parallelism, and perpendicularity information to construct line and plane equations, then check that each equation satisfies the original conditions.

Beginner 9 min read

What You'll Learn

  • Choose a point and direction vector to construct a line.
  • Construct a plane through a line and an additional point.
  • Use a cross product to find a direction satisfying two perpendicularity conditions.
  • Identify when geometric constraints determine a unique construction.
  • Verify constructed equations by substitution and dot products.
  • Recognize why a tempting but incorrect plane normal fails.

Turning Geometric Conditions into Equations

A line or plane is often described by conditions rather than by an equation. You might be told that a line passes through a point and is parallel to a given direction, or that a plane contains a line and another point. A reliable construction starts by translating each condition into information about points, directions, or normal vectors.

The basic checks are simple but important. A line contains a point when that point satisfies its equation. A direction lies along a line or within a plane when it is perpendicular to the plane's normal vector. A line is perpendicular to a plane when its direction is parallel to the plane's normal.

Construction principle: First identify a point on the desired object and the direction or normal that its conditions require. Then write an equation and test it against every given condition.

Equations Used in Constructions

A line through the point \(P=(x_0,y_0,z_0)\), with nonzero direction vector \(\mathbf{d}=(a,b,c)\), consists of points whose displacement from \(P\) is a scalar multiple of \(\mathbf{d}\). Its vector equation is \(\mathbf{r}=P+t\mathbf{d}\), where \(t\) is a real parameter. In coordinates, this means \(x=x_0+ta\), \(y=y_0+tb\), and \(z=z_0+tc\).

A plane through \(P=(x_0,y_0,z_0)\), with nonzero normal vector \(\mathbf{n}=(A,B,C)\), consists of points whose displacement from \(P\) is perpendicular to \(\mathbf{n}\). Thus its point-normal equation is \(\mathbf{n}\cdot((x,y,z)-P)=0\), or \(A(x-x_0)+B(y-y_0)+C(z-z_0)=0\).

For example, if a plane contains a direction \(\mathbf{d}\), its normal must satisfy \(\mathbf{n}\cdot\mathbf{d}=0\). If a line is perpendicular to a plane, its direction is parallel to that plane's normal. These statements turn geometric language into equations that can be used in a construction.

Worked Example: A Line Parallel to a Given Line

Worked Example: A Line Parallel to a Given Line

Find the line through \(P=(2,-1,4)\) parallel to the line through \(A=(1,0,2)\) and \(B=(3,2,1)\). First calculate a direction for the given line by subtracting its points:

$$ \overrightarrow{AB}=B-A=(3-1,\;2-0,\;1-2)=(2,2,-1). $$

The required line has the same direction and passes through \(P\). Therefore one equation is

$$ (x,y,z)=(2,-1,4)+t(2,2,-1), \qquad t\in\mathbb{R}. $$

In component form, \(x=2+2t\), \(y=-1+2t\), and \(z=4-t\). At \(t=0\), the line gives \((2,-1,4)\), as required. Its direction vector is \((2,2,-1)\), exactly the direction from \(A\) to \(B\), so it is parallel to the given line.

Constructing a Plane from a Line and a Point

Suppose a plane must contain a line with point \(A\) and direction \(\mathbf{d}\), and must also contain a point \(P\) not on that line. The vector \(P-A\) is another direction in the plane. A normal vector must be perpendicular to both directions, so a cross product provides one: \(\mathbf{n}=\mathbf{d}\times(P-A)\).

This construction gives a unique plane when \(\mathbf{d}\) and \(P-A\) are not parallel. If they are parallel, then \(P\) lies on the given line; the line and point do not determine a unique plane, because infinitely many planes can contain that line.

Worked Example: A Plane Containing a Line and a Point

Worked Example: A Plane Containing a Line and a Point

Find the plane containing the line through \(A=(1,0,2)\) with direction \(\mathbf{d}=(2,1,-1)\), and the point \(P=(0,2,1)\). The additional direction from \(A\) to \(P\) is

$$ P-A=(0-1,\;2-0,\;1-2)=(-1,2,-1). $$

Take the cross product to obtain a normal. Showing each component gives

$$ \begin{aligned} \mathbf{n} &=(2,1,-1)\times(-1,2,-1)\\ &=(1(-1)-(-1)(2),\;(-1)(-1)-2(-1),\;2(2)-1(-1))\\ &=(1,3,5). \end{aligned} $$

Use this normal and the point \(A\) in the point-normal equation:

$$ (x-1)+3(y-0)+5(z-2)=0, $$

so the plane is \(x+3y+5z-11=0\). Check the additional point: \(0+3(2)+5(1)-11=0\). Check that the line's direction lies in the plane: \((1,3,5)\cdot(2,1,-1)=2+3-5=0\). Finally, the line's specified point satisfies the plane equation: \(1+3(0)+5(2)-11=0\). Both the point and every point moving from it in direction \(\mathbf{d}\) are therefore in the plane.

Why the Cross Product Works

For two vectors \(\mathbf{u}=(u_1,u_2,u_3)\) and \(\mathbf{v}=(v_1,v_2,v_3)\), the cross product is

$$ \mathbf{u}\times\mathbf{v} =(u_2v_3-u_3v_2,\;u_3v_1-u_1v_3,\;u_1v_2-u_2v_1). $$

Its components are arranged so that it is perpendicular to both input vectors. For instance, taking its dot product with \(\mathbf{u}\) yields

$$ \begin{aligned} \mathbf{u}\cdot(\mathbf{u}\times\mathbf{v}) ={}&u_1(u_2v_3-u_3v_2) +u_2(u_3v_1-u_1v_3)\\ &+u_3(u_1v_2-u_2v_1)=0. \end{aligned} $$

The terms cancel in pairs: \(u_1u_2v_3\) cancels with \(-u_2u_1v_3\), \(-u_1u_3v_2\) cancels with \(u_3u_1v_2\), and \(u_2u_3v_1\) cancels with \(-u_3u_2v_1\). The same calculation with \(\mathbf{v}\cdot(\mathbf{u}\times\mathbf{v})\) also gives zero. Thus, when the cross product is nonzero, it supplies a direction perpendicular to both vectors. In plane construction, that is exactly the normal needed for two known directions within the plane.

Worked Example: A Line in a Plane with a Perpendicularity Condition

Worked Example: A Line in a Plane with a Perpendicularity Condition

Find a line through \(P=(1,1,2)\) that lies in the plane \(2x-y+2z=5\) and whose direction is perpendicular to \(\mathbf{v}=(1,2,0)\). The plane's normal is \(\mathbf{n}=(2,-1,2)\). A direction in the plane must be perpendicular to \(\mathbf{n}\); the required direction must also be perpendicular to \(\mathbf{v}\). Choose their cross product:

$$ \begin{aligned} \mathbf{d} &=\mathbf{n}\times\mathbf{v}\\ &=(2,-1,2)\times(1,2,0)\\ &=((-1)(0)-2(2),\;2(1)-2(0),\;2(2)-(-1)(1))\\ &=(-4,2,5). \end{aligned} $$

Thus a line with the required point and direction is

$$ (x,y,z)=(1,1,2)+t(-4,2,5), \qquad t\in\mathbb{R}. $$

Check that the point is in the plane: \(2(1)-1+2(2)=5\). Check that the direction stays in the plane: \(\mathbf{n}\cdot\mathbf{d}=2(-4)-2+2(5)=-8-2+10=0\). Hence every point on the line satisfies the plane equation. Also, \(\mathbf{d}\cdot\mathbf{v}=(-4)(1)+2(2)+5(0)=-4+4+0=0\), so the line's direction is perpendicular to \(\mathbf{v}\), as required. The cross product is nonzero, so it gives a valid direction.

Checking Compatibility and Avoiding a Common Error

Not every collection of conditions can be satisfied. For example, a plane containing a specified line and passing through a point not on that line is determined by those data. But if the point is on the line, the data do not determine a unique plane. Similarly, a proposed direction for a line in a plane must be perpendicular to the plane's normal; if its dot product with that normal is nonzero, it cannot lie in the plane.

A common error is to use the line's direction as the plane's normal. In the previous plane example, this would give \(\mathbf{n}=(2,1,-1)\). Although it produces a plane through \(A\) by writing \(2(x-1)+y-(z-2)=0\), that plane does not contain the specified line: its proposed normal dotted with the line direction is \(2(2)+1(1)+(-1)(-1)=6\), not zero. A plane containing the line needs a normal perpendicular to the line direction. The correct normal \((1,3,5)\) passes that test, since \(1(2)+3(1)+5(-1)=0\).

Verification checklist: Substitute each required point into the constructed equation. For a plane, dot its normal with every direction that should lie in the plane. For perpendicularity, check the relevant dot product. These checks test the original conditions, not merely the algebraic form of the answer.

Check Your Understanding

For each question, identify the required point and direction or normal before writing an equation.

  1. How do you find a direction vector for a line through two specified points?
  2. What condition must a plane's normal satisfy for a given direction to lie in the plane?
  3. Why does the cross product of two nonparallel directions give a normal to the plane they span?
  4. What geometric information is missing if a point specified for a plane lies on its given line?
  5. What dot product would you calculate to check whether a proposed line direction lies in a plane?

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