Straight lines in 3D

At a Glance

• Frequency: 5 sub-parts across 5 of 13 years (2015, 2018, 2019, 2021, 2023)
• Priority tier: T2
• Marks (count): 10 (2), 15 (2), 7 (1)
• Average solve time: ~10 min
• Difficulty mix: medium 2, easy 2, hard 1
• Section: A | Dominant type: computation

Why This Chapter Matters

Straight lines in 3D appear in every Section A (5-mark and 10-mark slots) as well as 15-mark compulsories, with at least one question in five of the last eleven years. The three archetypes — coplanarity/intersection, projection on a plane, and locus of a moving line — are each solved by a clean, repeatable algorithm. Once you own the scalar-triple-product coplanarity test and the projecting-plane construction, the first four questions become nearly mechanical; the 2023 locus proof requires parametric reasoning but follows a fixed strategy.

Minimum Theory

Three representations of a line. A line through the point $P_0 = (x_0, y_0, z_0)$ with direction vector $\mathbf{d} = (l, m, n)$ can be written in three equivalent forms:

• Vector form: $\mathbf{r} = \mathbf{P}_0 + t\,\mathbf{d}$, $t \in \mathbb{R}$.
• Parametric form: $x = x_0 + lt$, $y = y_0 + mt$, $z = z_0 + nt$.
• Symmetric (Cartesian) form: $\dfrac{x - x_0}{l} = \dfrac{y - y_0}{m} = \dfrac{z - z_0}{n}$.

To read direction ratios directly from the symmetric form, the denominator under each variable is one component of $\mathbf{d}$.

Coplanarity test. Two lines — $L_1$ through $P_1$ with direction $\mathbf{d}_1$, and $L_2$ through $P_2$ with direction $\mathbf{d}_2$ — are coplanar if and only if

$[\mathbf{P_2 - P_1},\; \mathbf{d}_1,\; \mathbf{d}_2] = 0,$

where $[\,\cdot\,,\,\cdot\,,\,\cdot\,]$ denotes the scalar triple product (determinant of the $3\times 3$ matrix formed by the three vectors as rows). Geometrically this says the displacement $\overrightarrow{P_1 P_2}$ lies in the plane spanned by $\mathbf{d}_1$ and $\mathbf{d}_2$.

Plane containing two coplanar lines. If $L_1$ and $L_2$ are coplanar, their common plane has normal

$\mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2$

and passes through $P_1$. The plane equation is $\mathbf{n} \cdot (\mathbf{r} - P_1) = 0$.

Projection of a line on a plane. The projection of line $L$ (direction $\mathbf{d}_L$) onto plane $\pi$ (normal $\mathbf{n}_\pi$) is found via the projecting plane $\pi'$: the unique plane containing $L$ and perpendicular to $\pi$. Its normal is $\mathbf{d}_L \times \mathbf{n}_\pi$. The projection is the line of intersection $\pi \cap \pi'$.

Question Archetypes

coplanar-lines-and-plane (3 question(s); 2015, 2019, 2021)

Test two lines for coplanarity/intersection and find the containing plane and intersection point

Recognition Cues

• The question gives two lines in symmetric or vector form and asks to “show they intersect”, “prove coplanarity”, “find the plane containing them”, or “find their point of intersection.”
• One line may be given by a point and direction with coordinates in arithmetic progression (AP) — this signals the AP-structure result.
• After coplanarity is confirmed, a second sub-task usually asks for the plane equation or the intersection point.

Solution Template

1. Read off $P_1$, $\mathbf{d}_1$ from $L_1$ and $P_2$, $\mathbf{d}_2$ from $L_2$.
2. Form $\mathbf{v} = P_2 - P_1$.
3. Evaluate the scalar triple product $[\mathbf{v}, \mathbf{d}_1, \mathbf{d}_2]$ as a $3\times 3$ determinant. If it equals zero, the lines are coplanar (or one line contains the other if also $\mathbf{d}_1 \times \mathbf{d}_2 = \mathbf{0}$).
4. Plane: Compute $\mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2$; simplify to lowest integer ratios. Write $\mathbf{n} \cdot (\mathbf{r} - P_1) = 0$.
5. Intersection: Write $P_1 + t\,\mathbf{d}_1 = P_2 + s\,\mathbf{d}_2$. Solve any two coordinate equations for $t$ and $s$, then verify the third equation is satisfied. Substitute back to find the point.

Worked Example 1

2015 Paper 1, 2015-P1-Q3c-ii (7 marks)

$L_1$ passes through $(a-d,\, a,\, a+d)$ with direction ratios $(\alpha-\delta,\, \alpha,\, \alpha+\delta)$; $L_2$ passes through $(b-c,\, b,\, b+c)$ with direction ratios $(\beta-\gamma,\, \beta,\, \beta+\gamma)$. Show $L_1$ and $L_2$ are coplanar and find their common plane.

Step 1 — Form the determinant.

$[\mathbf{P_2-P_1}, \mathbf{d}_1, \mathbf{d}_2] = \begin{vmatrix} (b-c)-(a-d) & b-a & (b+c)-(a+d)\\ \alpha-\delta & \alpha & \alpha+\delta\\ \beta-\gamma & \beta & \beta+\gamma \end{vmatrix}.$

Step 2 — Apply $C_1 + C_3 - 2C_2 = 0$.

For any row $(u-v,\, u,\, u+v)$, the combination $C_1 + C_3 - 2C_2 = (u-v)+(u+v)-2u = 0$. All three rows have this AP structure, so $C_1 + C_3 - 2C_2$ is the zero column. Hence:

$[\mathbf{P_2-P_1}, \mathbf{d}_1, \mathbf{d}_2] = 0. \qquad \checkmark$

Step 3 — Find the normal.

$\mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2 = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\ \alpha-\delta&\alpha&\alpha+\delta\\ \beta-\gamma&\beta&\beta+\gamma\end{vmatrix}.$

Computing: $n_x = \alpha(\beta+\gamma) - (\alpha+\delta)\beta = \alpha\gamma - \beta\delta,$

$n_y = -\bigl[(\alpha-\delta)(\beta+\gamma) - (\alpha+\delta)(\beta-\gamma)\bigr] = -2(\alpha\gamma-\beta\delta),$

$n_z = (\alpha-\delta)\beta - \alpha(\beta-\gamma) = \alpha\gamma-\beta\delta.$

So $\mathbf{n} = (\alpha\gamma-\beta\delta)(1,-2,1)$.

Step 4 — Plane through $P_1 = (a-d,a,a+d)$:

$1\cdot(x-(a-d)) - 2(y-a) + 1\cdot(z-(a+d)) = 0,$

$x - 2y + z - (a-d) + 2a - (a+d) = 0,$

$\boxed{x - 2y + z = 0.}$

The plane equation is independent of all six parameters — a consequence of the pure AP structure.

Worked Example 2

2019 Paper 1, 2019-P1-Q1e (10 marks)

Show that the lines $L_1: \mathbf{r} = (-1,3,-2) + t(-3,2,1)$ and $L_2: \mathbf{r} = (0,7,-7) + s(1,-3,2)$ intersect. Find the point of intersection and the plane containing them.

Step 1 — Coplanarity check.

$P_1 = (-1,3,-2)$, $\mathbf{d}_1 = (-3,2,1)$; $P_2 = (0,7,-7)$, $\mathbf{d}_2 = (1,-3,2)$.

$\mathbf{P_2-P_1} = (1,4,-5).$

$\begin{vmatrix}1&4&-5\\-3&2&1\\1&-3&2\end{vmatrix} = 1(4+3) - 4(-6-1) + (-5)(9-2) = 7 + 28 - 35 = 0. \qquad \checkmark$

Step 2 — Find $t$ and $s$.

$x$: $-1-3t = s \Rightarrow s = -1-3t$. $y$: $3+2t = 7-3s \Rightarrow 3+2t = 7-3(-1-3t) = 10+9t \Rightarrow 7t = -7 \Rightarrow t = -1$. Then $s = -1-3(-1) = 2$.

$z$-check: $-2+1(-1) = -3$ and $-7+2(2) = -3$. $\checkmark$

Step 3 — Intersection point.

$\mathbf{r} = (-1,3,-2) + (-1)(-3,2,1) = (2,1,-3).$

Step 4 — Containing plane.

$\mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2 = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\-3&2&1\\1&-3&2\end{vmatrix} = (4+3)\,\mathbf{i} - (-6-1)\,\mathbf{j} + (9-2)\,\mathbf{k} = (7,7,7) \propto (1,1,1).$

Plane through $(2,1,-3)$: $(x-2)+(y-1)+(z+3)=0$, i.e.\ $\boxed{x+y+z = 0}$.

Worked Example 3

2021 Paper 1, 2021-P1-Q4c (15 marks)

Find the plane containing the lines $L_1: (-1,-3,-5)+t(3,5,7)$ and $L_2: (2,4,6)+s(1,3,5)$. Also find their point of intersection.

Step 1 — Coplanarity.

$P_1=(-1,-3,-5)$, $\mathbf{d}_1=(3,5,7)$; $P_2=(2,4,6)$, $\mathbf{d}_2=(1,3,5)$.

$\mathbf{P_2-P_1} = (3,7,11).$

$\begin{vmatrix}3&7&11\\3&5&7\\1&3&5\end{vmatrix} = 3(25-21)-7(15-7)+11(9-5) = 12 - 56 + 44 = 0. \qquad \checkmark$

Step 2 — Normal and plane.

$\mathbf{n} = \mathbf{d}_1\times\mathbf{d}_2 = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\3&5&7\\1&3&5\end{vmatrix} = (25-21)\,\mathbf{i}-(15-7)\,\mathbf{j}+(9-5)\,\mathbf{k} = (4,-8,4) \propto (1,-2,1).$

Plane through $P_1=(-1,-3,-5)$: $(x+1)-2(y+3)+(z+5)=0 \implies \boxed{x-2y+z = 0.}$

Step 3 — Intersection point.

Equate components: $-1+3t = 2+s$, $-3+5t = 4+3s$, $-5+7t = 6+5s$.

From the first: $s = -3+3t$. Substitute into the second: $-3+5t = 4+3(-3+3t) = -5+9t \Rightarrow 2 = 4t \Rightarrow t = \tfrac{1}{2}$. Then $s = -3+\tfrac{3}{2} = -\tfrac{3}{2}$.

$z$-check: $-5+\tfrac{7}{2} = -\tfrac{3}{2}$ and $6+5(-\tfrac{3}{2}) = -\tfrac{3}{2}$. $\checkmark$

Intersection: $P_1 + \tfrac{1}{2}\mathbf{d}_1 = (-1+\tfrac{3}{2},\,-3+\tfrac{5}{2},\,-5+\tfrac{7}{2}) = \left(\tfrac{1}{2},-\tfrac{1}{2},-\tfrac{3}{2}\right)$.

$\boxed{\text{Intersection} = \left(\tfrac{1}{2},\,-\tfrac{1}{2},\,-\tfrac{3}{2}\right).}$

Common Traps

• The AP-structure plane is always $x-2y+z=0$: The parameters $a,b,c,d,\alpha,\beta,\gamma,\delta$ all cancel out; do not try to express the plane in terms of them. If your plane still contains these letters, recheck the column-combination step.
• Use different parameter names for the two lines ($t$ and $s$, never the same symbol). After solving two coordinate equations, always substitute into the third to confirm consistency — this is the check that distinguishes intersecting from skew.
• Cross product vs. scalar triple product: Coplanarity uses the scalar triple product (a number); the plane normal uses the cross product (a vector). Do not confuse them.
• Simplify the normal: $(4,-8,4)$ simplifies to $(1,-2,1)$. Using the unsimplified normal gives a correct plane but with a messy constant; simplifying first is cleaner.

line-projection-on-plane (1 question(s); 2018)

Project a line onto a plane via the projecting plane (contains line, perpendicular to plane)

Recognition Cues

• The question gives a line in symmetric form and a plane, and asks for “the projection of the line on the plane” or “the image of the line in the plane.”
• The answer is a line expressed as the intersection of two planes, or in symmetric form — never a single plane equation.

Solution Template

1. Read off the line point $P$ and direction $\mathbf{d}_L$ from the symmetric form; read off the plane normal $\mathbf{n}_\pi$.
2. Compute the projecting plane normal: $\mathbf{n}' = \mathbf{d}_L \times \mathbf{n}_\pi$.
3. Write the projecting plane $\pi'$: normal $\mathbf{n}'$, passes through $P$.
4. The projection is $\pi \cap \pi'$: give both plane equations as the answer.
5. Optionally find the direction of the projected line: $\mathbf{n}_\pi \times \mathbf{n}'$.

Worked Example

2018 Paper 1, 2018-P1-Q1e (10 marks)

Find the projection of the line $\dfrac{x-1}{2} = \dfrac{y-1}{3} = \dfrac{z+1}{-1}$ on the plane $x+y+2z = 6$.

Step 1 — Read data.

Line: point $P=(1,1,-1)$, direction $\mathbf{d}_L = (2,3,-1)$. Plane: normal $\mathbf{n}_\pi = (1,1,2)$.

Step 2 — Projecting plane normal.

$\mathbf{n}' = \mathbf{d}_L \times \mathbf{n}_\pi = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\2&3&-1\\1&1&2\end{vmatrix} = (6+1)\,\mathbf{i}-(4+1)\,\mathbf{j}+(2-3)\,\mathbf{k} = (7,-5,-1).$

Step 3 — Projecting plane $\pi'$ through $P=(1,1,-1)$:

$7(x-1)-5(y-1)-(z+1)=0 \implies \boxed{7x-5y-z = 3.}$

Step 4 — Projection line (intersection of the two planes):

$\boxed{\pi \cap \pi':\quad x+y+2z = 6, \quad 7x-5y-z = 3.}$

Step 5 — Direction of the projected line (optional but recommended):

$\mathbf{n}_\pi \times \mathbf{n}' = (1,1,2)\times(7,-5,-1) = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&1&2\\7&-5&-1\end{vmatrix} = (-1+10)\,\mathbf{i}-(−1-14)\,\mathbf{j}+(-5-7)\,\mathbf{k} = (9,15,-12) \propto (3,5,-4).$

Common Traps

• Order of the cross product matters: The projecting plane normal is $\mathbf{d}_L \times \mathbf{n}_\pi$, not $\mathbf{n}_\pi \times \mathbf{d}_L$. Either gives a valid normal to $\pi'$ (they are antiparallel), so the plane equation is the same — but be consistent.
• Give the answer as two plane equations, not one. A single plane does not define a line in 3D.
• Verify $\pi'$ contains the original line: Check that $P=(1,1,-1)$ satisfies $7x-5y-z=3$: $7-5+1=3$. $\checkmark$ Also check that $\mathbf{d}_L \cdot \mathbf{n}' = (2)(7)+(3)(-5)+(-1)(-1)=14-15+1=0$. $\checkmark$ (The line is parallel to $\pi'$, as required.)
• Verify $\pi' \perp \pi$: $\mathbf{n}_\pi \cdot \mathbf{n}' = (1)(7)+(1)(-5)+(2)(-1)=0$. $\checkmark$

locus-of-moving-line (1 question(s); 2023)

Find the locus traced by a line meeting given curves under a constraint

Recognition Cues

• The question describes a line meeting two or three specified guide curves/lines and asks to “prove the locus is” or “find the equation of the locus.”
• Guide curves are often: two skew lines given as intersections of planes (e.g.\ $y=mx, z=c$), and a circle or conic in a coordinate plane.
• The answer is a surface equation involving $x, y, z$ (typically a ruled quadric).

Solution Template

1. Parametrise one point on each guide curve using distinct parameters (e.g.\ $\alpha$, $\beta$).
2. Write the general point on the moving line as $P_1 + \lambda(P_2 - P_1)$ for parameter $\lambda$.
3. Impose the third constraint (e.g.\ lies on a circle) to get one equation relating the parameters.
4. Express $x$ and $y$ (or $x$, $y$, $z$) in terms of the parameters.
5. Eliminate all free parameters between the constraint equation and the coordinate expressions.
6. The resulting equation in $x$, $y$, $z$ is the locus.

Worked Example

2023 Paper 1, 2023-P1-Q4c (15 marks)

A variable line meets the lines $\ell_1: y=mx,\, z=c$ and $\ell_2: y=-mx,\, z=-c$ and also meets the circle $x^2+y^2=a^2,\, z=0$. Prove that the locus of the variable line is

$c^2m^2(cy-mzx)^2 + c^2(yz-cmx)^2 = a^2m^2(z^2-c^2)^2.$

Step 1 — Parametrise the three guide objects.

Point $A$ on $\ell_1$ (where $y=mx$, $z=c$): $A = (\alpha, m\alpha, c)$ for some $\alpha \in \mathbb{R}$. Point $B$ on $\ell_2$ (where $y=-mx$, $z=-c$): $B = (\beta, -m\beta, -c)$ for some $\beta \in \mathbb{R}$.

The variable line passes through $A$ and $B$. Its parametric form is: $\mathbf{r}(\lambda) = (1-\lambda)A + \lambda B = \bigl((1-\lambda)\alpha + \lambda\beta,\; (1-\lambda)m\alpha - \lambda m\beta,\; (1-2\lambda)c\bigr).$

Step 2 — Find the point on the circle ($z=0$).

Set $z = 0$: $(1-2\lambda)c = 0 \Rightarrow \lambda = \tfrac{1}{2}$.

At $\lambda = \tfrac{1}{2}$: $x_0 = \frac{\alpha+\beta}{2}, \qquad y_0 = \frac{m\alpha - m\beta}{2} = \frac{m(\alpha-\beta)}{2}.$

Step 3 — Apply the circle constraint.

$x_0^2 + y_0^2 = a^2$: $\frac{(\alpha+\beta)^2}{4} + \frac{m^2(\alpha-\beta)^2}{4} = a^2.$

$(\alpha+\beta)^2 + m^2(\alpha-\beta)^2 = 4a^2. \qquad(\ast)$

Step 4 — Express $\alpha$ and $\beta$ in terms of a general point $(x,y,z)$ on the moving line.

A general point on the line $AB$ at parameter $\lambda$ satisfies: $x = (1-\lambda)\alpha + \lambda\beta, \quad y = m(1-\lambda)\alpha - m\lambda\beta, \quad z = (1-2\lambda)c.$

From the $z$-equation: $\lambda = \tfrac{c-z}{2c}$, so $1-\lambda = \tfrac{c+z}{2c}$.

Then: $x = \frac{(c+z)\alpha + (c-z)\beta}{2c}, \qquad y = \frac{m(c+z)\alpha - m(c-z)\beta}{2c}.$

Solve for $\alpha$ and $\beta$. Add and subtract suitably: $2cx = (c+z)\alpha + (c-z)\beta, \qquad (1)$

$2cy = m(c+z)\alpha - m(c-z)\beta. \qquad (2)$

Multiply (1) by $m$: $2cmx = m(c+z)\alpha + m(c-z)\beta$. Add to (2): $2cmx + 2cy = 2m(c+z)\alpha \implies \alpha = \frac{c(mx+y)}{m(c+z)}.$

Multiply (1) by $m$ and subtract (2): $2cmx - 2cy = 2m(c-z)\beta \implies \beta = \frac{c(mx-y)}{m(c-z)}.$

Step 5 — Compute $\alpha + \beta$ and $\alpha - \beta$.

$\alpha + \beta = \frac{c(mx+y)}{m(c+z)} + \frac{c(mx-y)}{m(c-z)} = \frac{c}{m}\cdot\frac{(mx+y)(c-z)+(mx-y)(c+z)}{c^2-z^2}.$

Expand the numerator: $(mx+y)(c-z)+(mx-y)(c+z) = 2cmx - 2yz.$

So: $\alpha+\beta = \frac{c}{m}\cdot\frac{2(cmx-yz)}{c^2-z^2} = \frac{2c(cmx-yz)}{m(c^2-z^2)}.$

Similarly: $\alpha - \beta = \frac{c(mx+y)}{m(c+z)} - \frac{c(mx-y)}{m(c-z)} = \frac{c}{m}\cdot\frac{(mx+y)(c-z)-(mx-y)(c+z)}{c^2-z^2}.$

Expand: $(mx+y)(c-z)-(mx-y)(c+z) = 2cy - 2mxz.$

So: $\alpha-\beta = \frac{c}{m}\cdot\frac{2(cy-mzx)}{c^2-z^2} = \frac{2c(cy-mzx)}{m(c^2-z^2)}.$

Step 6 — Substitute into constraint $(*)$.

$\left(\frac{2c(cmx-yz)}{m(c^2-z^2)}\right)^2 + m^2\left(\frac{2c(cy-mzx)}{m(c^2-z^2)}\right)^2 = 4a^2.$

$\frac{4c^2(cmx-yz)^2}{m^2(c^2-z^2)^2} + \frac{4c^2(cy-mzx)^2}{(c^2-z^2)^2} = 4a^2.$

Multiply through by $m^2(c^2-z^2)^2 / 4$:

$c^2(cmx-yz)^2 + c^2m^2(cy-mzx)^2 = a^2m^2(c^2-z^2)^2.$

Since $(c^2-z^2)^2 = (z^2-c^2)^2$ and $(cmx-yz)^2 = (yz-cmx)^2$:

$\boxed{c^2m^2(cy-mzx)^2 + c^2(yz-cmx)^2 = a^2m^2(z^2-c^2)^2.} \qquad \blacksquare$

Common Traps

• Label the midpoint trick carefully. The symmetry $z=c$ and $z=-c$ forces the intersection with $z=0$ to be the midpoint of $AB$ ($\lambda=\tfrac{1}{2}$). Identifying this shortcut early saves a parameter.
• Keep the $m$ factors. In the expressions for $\alpha$ and $\beta$, the denominator carries $m$; dropping it introduces an incorrect factor throughout.
• $(c^2-z^2)^2 = (z^2-c^2)^2$: These are equal as squares, so either sign convention matches the target equation. Do not waste time rewriting — just note they are the same.
• This is a 15-mark proof question: Show every algebraic step explicitly — parametrisation, the midpoint calculation, the circle constraint substitution, and the final simplification. Skipping even one step will lose marks.

Marks-Aware Writing

7-mark question (2015): Show the determinant, write the column-combination argument in one line ($C_1+C_3-2C_2=0$), compute the cross product for the normal, and state the plane. Four clear steps = full marks.

10-mark question (2018, 2019): Show the scalar triple product calculation with the determinant written out (not just the result). For the intersection, display the three coordinate equations, show which two you solved, and verify the third. State the plane equation in a box. About 6–8 lines of working.

15-mark question (2021, 2023):

• Coplanarity (2021): Same as above but you must also show the intersection computation with both parameters and the $z$-check. The plane derivation and intersection together need roughly 10–12 lines.
• Locus proof (2023): A proof must be complete. The six steps above (parametrise → midpoint → circle constraint → solve for $\alpha, \beta$ → substitute → simplify) each carry marks. Write the boxed final equation followed by QED ($\blacksquare$). Partial proofs that reach only the constraint $(*)$ typically earn 7–8 out of 15.

Practice Set

• 2024-P1-Q2c (20 m) — — coplanarity + intersection + plane; treat as a combined 2021-type question
• 2013-P1-Q1d (10 m) — — line meets two lines; find point and plane
• 2015-P1-Q2d (13 m) — — AP-structure variant; practise the column-combination argument
• 2021-P1-Q1e (10 m) — — projection of a line on a plane
• 2020-P1-Q2c (15 m) — — locus-type problem; use parametric construction
• 2020-P1-Q3c (15 m) — — coplanar lines; verify and find plane
• 2017-P1-Q1e (10 m) — — standard coplanarity + intersection
• 2016-P1-Q1e (10 m) — — symmetric-form lines; find common plane
• 2016-P1-Q4a (10 m) — — projection of a line on a plane
• 2025-P1-Q1e (10 m) — — coplanar or projection; classify and solve
• 2025-P1-Q2c-i (10 m) — — intersection + plane; standard template