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Points, Lines, and Planes: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Read the notation

For distinct points $M$ and $N$ , name the endpoint or endpoints of $\overrightarrow{MN}$ and $\overline{MN}$ . State which of these objects contains the other.
1. Hint 1
  The mark above the letters tells you whether an object stops.
2. Hint 2
  A ray starts at its first letter and continues through its second.
Answer

$\overrightarrow{MN}$ has endpoint $M$ ; $\overline{MN}$ has endpoints $M$ and $N$ . The ray contains the segment.

Full solution

The ray begins at $M$ and continues through and beyond $N$ , so it has only one endpoint.

The segment stops at both $M$ and $N$ .
Every point between them belongs to the ray, so the ray contains the segment.

Answer

$\overrightarrow{MN}$ has endpoint $M$ ; $\overline{MN}$ has endpoints $M$ and $N$ . The ray contains the segment.

Key idea

Read the mark and the order together to identify a geometric object.
Problem 2 Same direction

Distinct points $A$ , $B$ , $C$ , and $D$ occur in that order on one line.

Among $\overrightarrow{BA}$ , $\overrightarrow{BC}$ , and $\overrightarrow{BD}$ , identify equal rays and a pair of opposite rays.

The labeled points lie on one line in the stated order. Drawn spacing adds no length or equality information.

Text description of this figure

The labeled points lie on one line in the stated order. Drawn spacing adds no length or equality information.
1. Hint 1
  First compare the endpoints, then the directions.
2. Hint 2
  From $B$ , points $C$ and $D$ are on the same side, while $A$ is on the other side.
Answer

$\overrightarrow{BC}=\overrightarrow{BD}$ . $\overrightarrow{BA}$ is opposite to either of those rays.

Full solution

All three rays begin at $B$ .
The ray toward $C$ also passes through $D$ , so the two forward rays are equal.

The ray toward $A$ extends the other way on the same line.
Together with either forward ray, it forms the entire line.

Answer

$\overrightarrow{BC}=\overrightarrow{BD}$ . $\overrightarrow{BA}$ is opposite to either of those rays.

Key idea

A ray's second named point selects its direction, not its extent.
Problem 3 Choose valid plane names

Plane $\mathcal S$ contains distinct collinear points $D,E,F$ and a point $Z$ off their line.

Among the triples $DEF$ , $DEZ$ , $DFZ$ , and $EFZ$ , which name $\mathcal S$ uniquely?

Explain why the remaining triple does not.

Plane S contains the complete line through distinct D and F, and point Z off that line. Point E is also on the drawn line. The parallelogram represents only a patch of the unlimited plane.

Text description of this figure

Plane S contains the complete line through distinct D and F, and point Z off that line. Point E is also on the drawn line. The parallelogram represents only a patch of the unlimited plane.
1. Hint 1
  A plane name needs three noncollinear points.
2. Hint 2
  Compare the triple made entirely from the line with each triple containing two line points and $Z$ .
Answer

$DEZ$ , $DFZ$ , and $EFZ$ name $\mathcal S$ uniquely. The collinear triple $DEF$ does not.

Full solution

The points $D,E,F$ lie on one line, so their triple does not determine a unique plane.

Each remaining triple contains two distinct points on that line and the point $Z$ off it.
Each triple is therefore noncollinear.

All its points lie in $\mathcal S$ , so the unique-plane foundation makes each of $DEZ$ , $DFZ$ , and $EFZ$ a valid name for $\mathcal S$ .

Answer

$DEZ$ , $DFZ$ , and $EFZ$ name $\mathcal S$ uniquely. The collinear triple $DEF$ does not.

Key idea

Select a plane name by checking its points, not by counting three letters.
Problem 4 Containment beyond a segment

A plane contains distinct points $P$ and $Q$ .

Point $R$ lies on $\overleftrightarrow{PQ}$ beyond $Q$ .

Must the plane contain $R$ ?

Give the accepted foundation that supports your answer.
1. Hint 1
  Decide whether the foundation concerns a segment or an entire line.
2. Hint 2
  The line has two distinct points in the plane.
Answer

Yes. The line-in-plane foundation puts all of $\overleftrightarrow{PQ}$ , including $R$ , in the plane.

Full solution

The plane contains two distinct points of $\overleftrightarrow{PQ}$ .
The accepted containment foundation therefore puts the entire line in the plane.

The location of $R$ beyond $Q$ does not matter, because the foundation applies beyond the segment as well.

Answer

Yes. The line-in-plane foundation puts all of $\overleftrightarrow{PQ}$ , including $R$ , in the plane.

Key idea

Two distinct points of a line in a plane force the whole line into that plane.
Problem 5 Three planes through one line

Three pairwise distinct planes $\mathcal A$ , $\mathcal B$ , and $\mathcal C$ all contain line $t$ . Describe the intersection of each pair. Can any pair be parallel?
1. Hint 1
  Every pair shares the whole given line.
2. Hint 2
  Apply the intersection foundation to distinct planes that share points.
Answer

Each pair intersects in line $t$ . No pair is parallel.

Full solution

Planes $\mathcal A$ and $\mathcal B$ share line $t$ , so they intersect.
Their distinctness makes all their common points form one line $k$ .

Choose two distinct points $U,V$ of $t$ .
Both lie on $k$ too, so the unique-line foundation identifies $k$ as $t$ .

The same reasoning applies to the other two pairs.
Parallel planes have no common points, so no pair here is parallel.

Answer

Each pair intersects in line $t$ . No pair is parallel.

Key idea

Several distinct planes can contain the same line.
Problem 6 Relationships on a box

A rectangular box has front face $ABFE$ , back face $DCGH$ , and connecting edges $AD$ , $BC$ , $FG$ , and $EH$ .

Use the box facts that each face is a rectangle and opposite faces lie in parallel planes. Classify each pair of complete lines:

Pair A: $BF$ and $EH$ .

Pair B: $BC$ and $FG$ .

Pair C: $EH$ and $HG$ .

The front face is ABFE and the back face is DCGH. Connecting edges are AD, BC, FG and EH. Only hidden edges AD, DC and DH are dashed. Use the given rectangular-face and parallel-plane facts, rather than perspective appearance.

Text description of this figure

The front face is ABFE and the back face is DCGH. Connecting edges are AD, BC, FG and EH. Only hidden edges AD, DC and DH are dashed. Use the given rectangular-face and parallel-plane facts, rather than perspective appearance.
1. Hint 1
  Check which vertices a common plane would have to contain, and use the stated rectangular faces.
2. Hint 2
  For the first pair, compare the plane through $B,F,E$ with the location of $H$ . For the other pairs, examine opposite and adjacent edges of their shared faces.
Answer

$BF$ and $EH$ are skew; $BC$ and $FG$ are parallel; $EH$ and $HG$ are perpendicular.

Full solution

Pair A: Any plane containing $BF$ and $EH$ contains $B,F,E$ .
These are noncollinear vertices of the rectangular front face, so that face is the only candidate plane.

Point $H$ lies in the distinct parallel back-face plane, outside the front-face plane.
The only candidate fails to contain $H$ , so $BF$ and $EH$ are skew.

Pair B: Lines $BC$ and $FG$ are opposite edge lines of rectangular face $BCGF$ , so they are parallel.

Pair C: Lines $EH$ and $HG$ are adjacent edge lines of the rectangular top face.
They meet at the square corner $H$ , so they are perpendicular.

Answer

$BF$ and $EH$ are skew; $BC$ and $FG$ are parallel; $EH$ and $HG$ are perpendicular.

Key idea

Classify line pairs using incidence and the given relationships of a solid's faces.
Problem 7 Two planes containing two lines

Planes $\mathcal P$ and $\mathcal Q$ both contain two distinct intersecting lines $r$ and $s$ .

Must the planes coincide?

Explain using the incidence foundations.
1. Hint 1
  Find three noncollinear points that both planes must contain.
2. Hint 2
  Choose the lines' meeting point and one different point on each line.
Answer

Yes. The planes coincide.

Full solution

Let $V$ be the meeting point.
Choose $X$ on $r$ and $Y$ on $s$ , each different from $V$ .

Distinct lines already share $V$ and can have no other common point.
Thus $Y$ is off $r$ , the unique line through $V$ and $X$ , making $V,X,Y$ noncollinear.

Both planes contain the three points.
Exactly one plane contains that noncollinear triple, so $\mathcal P$ and $\mathcal Q$ are the same plane.

Answer

Yes. The planes coincide.

Key idea

Two distinct intersecting lines supply a noncollinear triple and determine one plane.
Problem 8 Disjoint does not settle it

A rectangular box has front face $ABFE$ , back face $DCGH$ , and connecting edges $AD$ , $BC$ , $FG$ , and $EH$ .

Each face is a rectangle, and opposite faces lie in parallel planes.

The claim is: any two lines with no common point are parallel.

Use a pair of the box's complete edge lines to give a counterexample, explain why it meets the premise, and identify the missing condition.

The front face is ABFE and the back face is DCGH. Connecting edges are AD, BC, FG and EH. Only hidden edges AD, DC and DH are dashed. Use the given rectangular-face and parallel-plane facts, rather than perspective appearance.

Text description of this figure

The front face is ABFE and the back face is DCGH. Connecting edges are AD, BC, FG and EH. Only hidden edges AD, DC and DH are dashed. Use the given rectangular-face and parallel-plane facts, rather than perspective appearance.
1. Hint 1
  A plane containing both selected lines would have to contain all their endpoints.
2. Hint 2
  Compare a front vertical edge with a back horizontal edge. Which face plane would three of their endpoints determine?
Answer

For example, lines $AE$ and $CD$ are skew. They have no common point but are not parallel; the claim is missing coplanarity.

Full solution

Take lines $AE$ and $CD$ .
Any common plane would contain the noncollinear points $A,C,D$ , so it would be the bottom-face plane.
But $E$ is in the parallel top-face plane, outside the bottom-face plane.
Thus the two lines are noncoplanar.

They are skew and have no common point, since noncoplanar lines cannot meet.
They satisfy the claim's premise but fail its parallelism conclusion.
Requiring the distinct lines to be coplanar makes the claim correct.

Answer

For example, lines $AE$ and $CD$ are skew. They have no common point but are not parallel; the claim is missing coplanarity.

Key idea

A self-contained counterexample satisfies the hypothesis while failing the conclusion.
Problem 9 A wall and a shelf

Model a wall and a shelf as distinct planes sharing a line.

A straight rail contains distinct points $R$ and $S$ on that shared line.

Must the rail's entire line belong to both planes?

Explain, and name the intersection line.
1. Hint 1
  The two distinct points $R$ and $S$ determine the rail's line.
2. Hint 2
  Apply line containment separately to the wall plane and the shelf plane.
Answer

Yes. Their intersection is $\overleftrightarrow{RS}$ , the rail's line, which lies in both planes.

Full solution

The rail's distinct points $R$ and $S$ lie on the shared intersection line.
Line uniqueness identifies both the rail's line and the intersection line as $\overleftrightarrow{RS}$ .

Both planes contain that entire line by the line-in-plane foundation.
This conclusion concerns the ideal line extending beyond the physical rail.

Answer

Yes. Their intersection is $\overleftrightarrow{RS}$ , the rail's line, which lies in both planes.

Key idea

Two shared distinct points can name the intersection line of distinct planes.
Problem 10 Enough information to decide

In each case, decide whether the information forces a unique plane.

Case A: three distinct points on one line.

Case B: a line and a point off that line.

Case C: two distinct intersecting lines.

Explain the decisive condition in each case.
1. Hint 1
  Try to locate three noncollinear points in each case.
2. Hint 2
  For Case C, choose the intersection and one other point from each line.
Answer

Case A: no. Case B: yes. Case C: yes.

Full solution

Case A supplies only collinear points.
Many planes contain their common line, so no unique plane is forced.

In Case B, choose distinct points $J,K$ on the given line and name the off-line point $T$ .
The given line is the unique line through $J,K$ , and $T$ is off it, so $J,K,T$ are noncollinear.

Their unique plane contains the entire given line by the line-in-plane foundation.
Any plane containing that line and $T$ contains the same noncollinear triple, so it must be this plane.

In Case C, let the distinct lines $r,s$ meet at $V$ .

Choose $X$ on $r$ and $Y$ on $s$ , each different from $V$ .

Distinct lines already share $V$ and can have no other common point, so $Y$ is off $r$ , the unique line through $V,X$ .

Thus $V,X,Y$ are noncollinear.

Their unique plane contains both lines by the line-in-plane foundation.
Any plane containing both lines contains the same noncollinear triple, so it must be this plane.

Case A shows three collinear points H, L and N. Case B shows points J and K on a line and T off it. Case C shows lines r and s meeting at V, with X on r and Y on s.

Answer

Case A: no. Case B: yes. Case C: yes.

Key idea

Sufficient spatial information often reduces to finding three noncollinear points.

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