CP 6.5 - Where a line meets a plane | Edexcel A-Level Further Mathematics

Substituting a line into a plane reduces a spatial intersection to one equation. You will find the intersection point, distinguish a parallel line from one contained in the plane, and determine constants from a specified intersection condition.

Finding the intersection point

Can a straight line meet a flat plane infinitely many times? Usually there is one crossing, but the equation will reveal the exceptional case too.

Begin with $\mathbf r=(1,2,3)^{\mathrm T}+t(2,-1,1)^{\mathrm T}$ . Its coordinates are $x=1+2t,y=2-t,z=3+t$ . On the plane $x+y+z=10$ , these must add to $10$ , so the parameter is found from a single linear equation.

You already know

A point on a line uses the same parameter in all three coordinates. A point belongs to $\mathbf r.\mathbf n=d$ exactly when its position vector satisfies that scalar-product equation.

The general substitution exposes the same structure.

Line-plane intersection equation

For line $\mathbf r=\mathbf a+t\mathbf b$ , $\mathbf b\ne\mathbf0$ , and plane $\mathbf r.\mathbf n=d$ , $\mathbf n\ne\mathbf0$ ,

\mathbf a.\mathbf n+t(\mathbf b.\mathbf n)=d

If $\mathbf b.\mathbf n\ne0$ , then

t=\frac{d-\mathbf a.\mathbf n}{\mathbf b.\mathbf n}

Substitute this parameter back into the line to obtain the point. Not in the formula booklet: learn it.

You can simply substitute the coordinate expressions rather than memorise the fraction. It is the same linear equation.

Worked example 1: a unique crossing

Question. Find where $\mathbf r=(1,2,3)^{\mathrm T}+t(2,-1,1)^{\mathrm T}$ meets $x+y+z=10$ .

Plan. Insert all three coordinate expressions into the plane, solve for $t$ , then return to the line.

\begin{aligned} (1+2t)+(2-t)+(3+t)&=10 && \text{point must satisfy the plane} \\ 6+2t&=10\quad\Longrightarrow\quad t=2 && \text{solve the linear equation} \\ (x,y,z)&=(1+4,2-2,3+2)=(5,0,5) && \text{substitute into the line} \end{aligned}

Why this step? $t=2$ identifies a location along the line; it is not itself a coordinate triple.

Answer. The intersection is $(5,0,5)$ .

Check. $5+0+5=10$ , so the point lies in the plane as well as on the line.

Worked example 2 changes only the plane constant, from $10$ to $14$ . Its orientation stays fixed while its intersection moves along the line.

Worked example 2: a parallel plane at a different position

Question. Find where the same line meets $x+y+z=14$ .

Plan. Update the right-hand side, then calculate the new point.

\begin{aligned} 6+2t&=14 && \text{same coordinate sum} \\ t&=4 && \text{solve} \\ (x,y,z)&=(1+8,2-4,3+4)=(9,-2,7) && \text{return to the line} \end{aligned}

Why this step? Keeping the old point would ignore the plane's changed position, even though its normal is unchanged.

Answer. $(9,-2,7)$ .

Check. $9-2+7=14$ , and the displacement from the previous intersection is $2(2,-1,1)$ , parallel to the line.

A line segment needs one further check: its parameter must lie in the segment's permitted interval. An infinite line may cross a plane beyond the end of a specified segment.

Watch out

Wrong: use different parameters in $x,y,z$ , or stop after finding $t$ . Right: substitute the same solved parameter into all three coordinates and state the point.

Move the plane to a third parallel position.

Parallel to a plane or contained in it

If the line direction is perpendicular to the normal, substitution removes the parameter completely, leaving either a true identity or a contradiction.

For example, $\mathbf r=(1,2,3)^{\mathrm T}+t(1,1,-2)^{\mathrm T}$ always satisfies

x+y+z=(1+t)+(2+t)+(3-2t)=6

It lies entirely in $x+y+z=6$ , but never meets $x+y+z=7$ . The direction is the same in both cases; the anchor decides which geometry occurs.

Substitution result	Geometric conclusion
A non-zero multiple of $t$ remains	One intersection point
A true statement independent of $t$	The whole line lies in the plane
A false statement independent of $t$	The line is parallel to, and separate from, the plane

[DIAGRAM: asset_name: The three line-plane intersection cases; asset_slug: 9fm0-cp43-intersection-cases; caption: A unique parameter, an identity and a contradiction describe three different geometries; description: Three exact cross-section diagrams show a clay line crossing a blue plane trace once, a clay line lying within the plane trace, and a clay line running parallel above it. The panels are labelled one point, contained, and separate and parallel, with corresponding labels one parameter value, every parameter value, and no parameter value]
A unique parameter, an identity and a contradiction describe three different geometries

The scalar-product form says the same thing. If $\mathbf b.\mathbf n=0$ , compare $\mathbf a.\mathbf n$ with $d$ . Equality means contained; inequality means separate. These classification consequences are not in the formula booklet; derive them from the substituted equation.

A line contained in a plane

A line is contained in a plane when every point of the line belongs to the plane. Its direction is parallel to the plane and one, hence every, anchor lies in it. A separate parallel line has the same directional condition but no points in the plane.

There is a useful surprise here: a line and a plane can have zero, one or infinitely many common points, but never exactly two. If two distinct line points satisfy a linear plane equation, every point obtained by continuing along their displacement satisfies it too.

Key idea

A zero direction-normal product tells you that the line cannot cross the plane transversely. Test an anchor to distinguish containment from separation.

The opening question has its answer: a line can meet a plane infinitely many times by lying in it. Check the anchor before concluding that there is no meeting.

Determining a constant from the intersection

A specified coordinate of the intersection can determine an unknown constant in the line, provided you use both the line and plane conditions.

For $\mathbf r=(2,1,k)^{\mathrm T}+t(1,-1,2)^{\mathrm T}$ , the first coordinate gives $x=2+t$ . Once a target value of $x$ fixes $t$ , the plane equation can determine $k$ . The unknown constant describes a family of lines; the running parameter locates a point on one line.

Worked example 3: prescribing the first coordinate

Question. The line $\mathbf r=(2,1,k)^{\mathrm T}+t(1,-1,2)^{\mathrm T}$ meets $x+y+z=8$ at a point with $x=4$ . Find $k$ .

Plan. Use the given first coordinate to find $t$ , then substitute the full point into the plane.

\begin{aligned} 2+t&=4\quad\Longrightarrow\quad t=2 && \text{specified coordinate} \\ (x,y,z)&=(4,-1,k+4) && \text{point on the line} \\ 4-1+k+4&=8\quad\Longrightarrow\quad k=1 && \text{point in the plane} \end{aligned}

Why this step? The coordinate condition fixes the parameter, while plane membership fixes the line's constant; they play different roles.

Answer. $k=1$ .

Check. The point is $(4,-1,5)$ , and $4-1+5=8$ .

Worked example 4 changes only the prescribed first coordinate, from $4$ to $5$ .

Worked example 4: selecting another member of the line family

Question. Keep the same line family and plane, but require the intersection to have $x=5$ . Find $k$ .

Plan. Recalculate the parameter and then the plane constraint.

\begin{aligned} 2+t&=5\quad\Longrightarrow\quad t=3 && \text{new first coordinate} \\ (x,y,z)&=(5,-2,k+6) && \text{same line family} \\ 5-2+k+6&=8\quad\Longrightarrow\quad k=-1 && \text{plane condition} \end{aligned}

Why this step? Changing the target point does not mean replacing the fixed parameter $k$ by the running parameter $t$ .

Answer. $k=-1$ .

Check. The point $(5,-2,5)$ satisfies the plane and the chosen line.

Here $x+y=3$ on every line in the family, so meeting $x+y+z=8$ always gives $z=5$ . This is a helpful check derived from the equations, not an extra memorised formula.

Watch out

Wrong: treat $k$ and $t$ as interchangeable. Right: $k$ selects which line you have; $t$ selects a point along it. Write the two conditions separately before solving.

Use a third prescribed coordinate to select another line.

Quick Recap

Substitute a line's coordinate expressions into the plane and solve for its parameter.
Return to the line to calculate the actual point.
A vanished parameter gives either a whole line in the plane or a separate parallel line; test the remaining statement.
Distinguish a fixed constant selecting a line from its running parameter selecting a point.

A line can meet a plane infinitely many times: the substituted equation must hold for every parameter, so the whole line lies in the plane.

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CP 6.5 - Where a line meets a plane

Finding the intersection point

Parallel to a plane or contained in it

Determining a constant from the intersection

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