Curves in space: tangent, normal, binormal

At a Glance

• Frequency: 2 sub-parts across 2 of 13 years (2013, 2018)
• Priority tier: T3
• Marks (count): 10 (2)
• Average solve time: ~6 min
• Difficulty mix: easy 2
• Section: B | Dominant type: proof

Why This Chapter Matters

Both UPSC questions on curves in space are compulsory 10-mark items that take under seven minutes each. The 2013 question asks you to show a parametric curve lies in a plane — solved in two lines by spotting a linear relation among the components. The 2018 question asks for the angle between a curve’s tangent and a fixed line — a direct dot-product calculation. Neither question requires curvature or torsion, just $\mathbf{r}'(t)$ and basic linear algebra. Mastering these two patterns secures 20 fast marks.

Minimum Theory

Parametric curves and tangent vectors. For a curve $\mathbf{r}(t) = (x(t), y(t), z(t))$, the tangent vector at $t$ is $\mathbf{r}'(t) = (\dot x, \dot y, \dot z)$. A unit tangent vector is $\hat{\mathbf{T}} = \mathbf{r}'/|\mathbf{r}'|$.

Angle between a curve and a line. If the curve’s tangent direction is $\mathbf{T} = \mathbf{r}'(t)$ and the line has direction $\mathbf{L}$, the angle $\theta$ between them satisfies

$\cos\theta = \frac{|\mathbf{T}\cdot\mathbf{L}|}{|\mathbf{T}||\mathbf{L}|}.$

The angle may depend on the parameter $t$ (the general point).

Planarity criterion. A curve $\mathbf{r}(t)$ lies in a plane iff there exists a constant normal $\hat{\mathbf{n}}$ and constant $d$ such that $\hat{\mathbf{n}}\cdot\mathbf{r}(t) = d$ for all $t$. Equivalently: eliminate $t$ algebraically and find a linear relation $ax + by + cz = d$. The quickest approach is to spot common expressions among the component formulas.

Direction of a line from two planes. The line of intersection of two planes $f_1(x,y,z)=0$ and $f_2(x,y,z)=0$ has direction $\mathbf{n}_1\times\mathbf{n}_2$ where $\mathbf{n}_i$ are the normals to the planes. For the special case ”$y=0$ and $z=x$”, a point on the line satisfies $y=0$ and $z=x$, so the direction is $(1,0,1)$.

Question Archetypes

curve-planarity (1 question(s); 2013)

Recognition Cues

• “Show the curve $\mathbf{r}(t)$ lies in a plane.”
• The components are rational or trigonometric in $t$; there is usually a hidden linear combination $ax + by + cz =$ constant.
• Faster than computing torsion $= 0$.

Solution Template

1. Write $x(t), y(t), z(t)$ explicitly.
2. Look for a linear combination $ax + by + cz$ that simplifies (rational expressions often share $1/t$ terms that cancel).
3. Derive the plane equation $ax + by + cz = d$ and verify it holds identically in $t$.
4. State: “the curve lies in the plane $ax + by + cz = d$.”

Worked Example

2013 Paper 1, 2013-P1-Q5e (10 marks)

Show that the curve $\mathbf{x}(t) = t\hat\imath + \dfrac{1+t}{t}\hat\jmath + \dfrac{1-t^2}{t}\hat k$ lies in a plane.

Step 1 — Components.

$x = t, \quad y = \frac{1+t}{t} = \frac{1}{t}+1, \quad z = \frac{1-t^2}{t} = \frac{1}{t}-t.$

Step 2 — Spot the relation. Both $y - 1$ and $z + x$ equal $1/t$:

$y - 1 = \frac{1}{t}, \qquad z + x = \frac{1}{t} + t - t = \frac{1}{t}.$

Wait: $z + x = \frac{1}{t} - t + t = \frac{1}{t}$. So $y - 1 = z + x$, i.e. $x - y + z + 1 = 0$.

Step 3 — Verify identically. For all $t \ne 0$:

$x - y + z + 1 = t - \!\left(\frac{1}{t}+1\right) + \!\left(\frac{1}{t}-t\right) + 1 = t - \frac{1}{t} - 1 + \frac{1}{t} - t + 1 = 0. \checkmark$

$\boxed{\text{The curve lies in the plane } x - y + z + 1 = 0.}$

Common Traps

• Computing torsion $\tau = 0$ also proves planarity, but it requires three derivatives and a triple product — far more work. Spot the plane algebraically first.
• The curve is undefined at $t = 0$; the plane statement applies for all $t \ne 0$.
• The verification (Step 3) must hold identically in $t$, not just for specific values — always check algebraically.

tangent-line-angle (1 question(s); 2018)

Recognition Cues

• “Find the angle between the tangent at a general point of the curve $\mathbf{r}(t)$ and the line $[\ldots]$.”
• The line is given as the intersection of two planes; find its direction by reading off the common solution.
• The angle typically depends on $t$ — leave as a formula.

Solution Template

1. Differentiate: $\mathbf{T} = \mathbf{r}'(t)$.
2. Find the direction $\mathbf{L}$ of the fixed line (from the two-plane description).
3. Compute $\cos\theta = (\mathbf{T}\cdot\mathbf{L})/(|\mathbf{T}||\mathbf{L}|)$.
4. State the angle as a function of $t$.

Worked Example

2018 Paper 1, 2018-P1-Q5b (10 marks)

Find the angle between the tangent at a general point of the curve $x=3t,\;y=3t^2,\;z=3t^3$ and the line $y=z-x=0$.

Step 1 — Tangent vector.

$\mathbf{r}'(t) = (3, 6t, 9t^2) \parallel (1, 2t, 3t^2).$

Step 2 — Direction of the line. "$y = z - x = 0$" means $y = 0$ and $z = x$. A point on the line is $(s, 0, s)$, so the direction is

$\mathbf{L} = (1, 0, 1).$

Step 3 — Angle.

$\mathbf{T}\cdot\mathbf{L} = 1\cdot 1 + 2t\cdot 0 + 3t^2\cdot 1 = 1+3t^2.$

$|\mathbf{T}| = \sqrt{1+4t^2+9t^4}, \qquad |\mathbf{L}| = \sqrt{2}.$

$\cos\theta = \frac{1+3t^2}{\sqrt{2}\sqrt{1+4t^2+9t^4}}.$

$\boxed{\theta = \cos^{-1}\!\left(\frac{1+3t^2}{\sqrt{2(1+4t^2+9t^4)}}\right).}$

At $t=0$: $\cos\theta = 1/\sqrt2$, so $\theta = 45°$. The angle varies with $t$.

Common Traps

• "$y = z - x = 0$" is two equations $y = 0$ and $z - x = 0$, not a vector normal to any plane. Read the direction from the parametric form $(s, 0, s)$.
• The angle depends on $t$ — there is no single “the angle.” Give the formula and optionally evaluate at $t=0$.
• Drop the common factor of $3$ from $\mathbf{r}'$ before computing dot products to reduce algebra.

Marks-Aware Writing

A 10-mark answer must: write down $\mathbf{r}'(t)$ or the component formulas clearly; execute the key step (spotting the plane equation, or computing the dot product); verify or simplify; state the final result. In the planarity problem, the verification step (showing the relation holds for all $t$) is worth 3–4 marks and must be algebraic, not numeric. In the angle problem, identifying the line direction correctly is worth 3 marks; the dot product calculation is the remaining 7.

Practice Set

• 2013-P1-Q8b (10 m) — Related space-curve problem. ()
• 2017-P1-Q5e (10 m) — Tangent to a space curve. ()
• 2017-P1-Q7a (16 m) — Space curve geometry. ()
• 2025-P1-Q6c-i (10 m) — Tangent/normal direction on a parametric curve. ()