Angles and intersections

Two lines can cross at an angle or miss each other. Two planes meet along an angle too, and so does a line slicing through a plane — and often you need the exact point where that piercing happens. This closing piece of Three-Dimensional Geometry gives you three angle formulas built from the same dot-product idea, plus a clean method for finding where a line meets a plane. The line-line and plane-plane formulas share the familiar cosine shape you already used with direction cosines. The line-plane case looks different — it uses sine, not cosine — because the natural first step is to find the angle with the plane's normal, then take the complement. Master when to use which, and this becomes one of the more mechanical, mark-scoring corners of the chapter.

The core insight

The angle between two lines or two planes uses $\cos\theta$ on direction ratios (or normals), but the angle between a line and a plane uses $\sin\phi$ because it is the complement of the angle the line makes with the plane's normal.

The core idea

Angle between two lines. For lines with direction ratios $a_1,b_1,c_1$ and $a_2,b_2,c_2$ ,

\cos\theta = \frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}.

where: $\theta$ is the acute angle between the lines. They are parallel when $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$ , and perpendicular when $a_1a_2+b_1b_2+c_1c_2=0$ .

Angle between two planes. Planes $a_1x+b_1y+c_1z+d_1=0$ and $a_2x+b_2y+c_2z+d_2=0$ have normals with direction ratios $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$ . The angle between the planes equals the angle between these normals, so the formula is algebraically the same shape as the line-line one, just applied to normal direction ratios instead of line direction ratios:

\cos\theta = \frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}.

Planes are parallel when their normal direction ratios are proportional, and perpendicular when $a_1a_2+b_1b_2+c_1c_2=0$ .

Angle between a line and a plane. For a line with direction ratios $a,b,c$ and a plane with normal direction ratios $a',b',c'$ , the angle $\phi$ between the line and the plane itself is

\sin\phi = \frac{|aa'+bb'+cc'|}{\sqrt{a^2+b^2+c^2}\sqrt{a'^2+b'^2+c'^2}}.

where: $\phi$ is measured between the line and the plane, not between the line and the normal — that is why sine, not cosine, appears. A line is parallel to the plane when $aa'+bb'+cc'=0$ ; a line is normal to the plane when its direction ratios are proportional to $a',b',c'$ .

Intersection of a line and a plane. To find where line $\dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=\dfrac{z-z_1}{c}=\lambda$ meets plane $px+qy+rz+s=0$ : write the line's general point as $(x_1+a\lambda,\ y_1+b\lambda,\ z_1+c\lambda)$ , substitute into the plane's equation, and solve the resulting linear equation in $\lambda$ . If $\lambda$ 's coefficient vanishes and the constant term is also zero, the line lies entirely in the plane; if the coefficient vanishes but the constant term does not, the line is parallel to the plane and never meets it.

Figure

A line piercing a tilted plane at one point with the normal drawn from that point and the acute angle phi marked between the line and the plane

Derivation AWhere the sine formula for a line and a plane comes from

The goal is to show why the angle between a line and a plane is found using sine of the dot product between direction ratios and normal direction ratios, rather than cosine, and why the absolute value sign is essential — a confusion that costs marks whenever the raw dot product turns out negative.

Step 1. Call $\psi$ the angle between the line (direction ratios $a,b,c$ ) and the plane's normal (direction ratios $a',b',c'$ ). Since these are two ordinary directions in space, the standard line-line cosine formula applies directly to them: $\cos\psi = \dfrac{aa'+bb'+cc'}{\sqrt{a^2+b^2+c^2}\sqrt{a'^2+b'^2+c'^2}}$ .

Step 2. The angle $\phi$ that the line makes with the plane itself is not $\psi$ . Picture the normal standing straight up out of the plane: if the line lies flat in the plane, $\psi=90^\circ$ and $\phi=0^\circ$ ; if the line lies along the normal, $\psi=0^\circ$ and $\phi=90^\circ$ . So $\phi$ and $\psi$ always add to $90^\circ$ , meaning $\phi = 90^\circ-\psi$ .

Step 3. Apply the co-function identity $\sin(90^\circ-\psi)=\cos\psi$ to get $\sin\phi=\cos\psi$ . Substituting Step 1's expression for $\cos\psi$ gives $\sin\phi = \dfrac{aa'+bb'+cc'}{\sqrt{a^2+b^2+c^2}\sqrt{a'^2+b'^2+c'^2}}$ .

Step 4. By convention $\phi$ is taken as the acute angle between $0^\circ$ and $90^\circ$ , so $\sin\phi$ must be non-negative even though the raw dot product $aa'+bb'+cc'$ can come out negative depending on how the direction ratios were written down. Wrapping the numerator in an absolute value fixes this without changing the geometry, giving the final formula $\sin\phi = \dfrac{|aa'+bb'+cc'|}{\sqrt{a^2+b^2+c^2}\sqrt{a'^2+b'^2+c'^2}}$ .

Worked example

Find the point of intersection of the line $\dfrac{x-2}{3}=\dfrac{y+1}{4}=\dfrac{z-2}{2}$ with the plane $2x+y+z=5$ , and find the angle this line makes with the plane $x-y+2z=7$ .

Step 1. Parametrise the line by setting $\dfrac{x-2}{3}=\dfrac{y+1}{4}=\dfrac{z-2}{2}=\lambda$ , giving the general point $x=2+3\lambda,\ y=-1+4\lambda,\ z=2+2\lambda$ .

Step 2. Substitute into $2x+y+z=5$ : $2(2+3\lambda)+(-1+4\lambda)+(2+2\lambda)=5$ , which expands to $4+6\lambda-1+4\lambda+2+2\lambda=5$ , so $5+12\lambda=5$ and $\lambda=0$ .

Step 3. At $\lambda=0$ , the point is $(x,y,z)=(2,-1,2)$ . Check: $2(2)+(-1)+2=5$ ✓, so the line meets the plane $2x+y+z=5$ at $(2,-1,2)$ — which is also the line's own base point.

Step 4. For the angle with plane $x-y+2z=7$ : the line's direction ratios are $(3,4,2)$ and the normal's are $(1,-1,2)$ , so $aa'+bb'+cc'=(3)(1)+(4)(-1)+(2)(2)=3-4+4=3$ . The magnitudes are $\sqrt{9+16+4}=\sqrt{29}$ and $\sqrt{1+1+4}=\sqrt6$ , so

\sin\phi = \frac{|3|}{\sqrt{29}\cdot\sqrt6} = \frac{3}{\sqrt{174}}.

The line meets $2x+y+z=5$ at the point $(2,-1,2)$ , and makes an angle $\phi=\sin^{-1}\!\left(\dfrac{3}{\sqrt{174}}\right)$ with the plane $x-y+2z=7$ .

Competitive stretch

For the earlier pair line $\dfrac{x-1}{2}=\dfrac{y+1}{2}=\dfrac{z}{1}$ and plane $3x+4y+12z=13$ : direction ratios $(2,2,1)$ , normal $(3,4,12)$ , dot product $6+8+12=26$ , magnitudes $3$ and $13$ , so $\sin\phi=\dfrac{26}{39}=\dfrac23$ . Since the dot product is nonzero, the line is not parallel to this plane — it meets it at $\phi=\sin^{-1}(2/3)$ .

Common traps

Using $\cos$ instead of $\sin$ for the line-plane angle — this case is genuinely different because the direct dot-product calculation gives the angle with the normal, not with the plane.
Inverting the parallel condition: a line is parallel to a plane when $aa'+bb'+cc'=0$ (line perpendicular to normal), the opposite pattern from line-parallel-to-line, and easy to get backwards under pressure.
Forgetting to check whether $\lambda$ 's coefficient vanishes when solving for an intersection — if it does, the line either lies in the plane (constant term also zero) or is strictly parallel to it (constant term nonzero), and there is no single intersection point.
Applying the plane-plane cosine formula to two lines that merely lie in the planes instead of to the planes' normals — the correct comparison is always normal-to-normal.

Ask the tutor: if you want to see this method applied to your own line and plane equations, ask the AI tutor to walk through it step by step.

Why does the line-plane formula use sine when line-line and plane-plane both use cosine?

Because the dot product of the line's direction ratios with the plane's normal direction ratios directly gives you the cosine of the angle between the line and the normal, not between the line and the plane. Since the line-and-plane angle is the complement of the line-and-normal angle, you convert using $\sin\phi=\cos(90^\circ-\phi)$ , and that complement is exactly what turns the formula into a sine. Line-line and plane-plane comparisons never need this complement step because you are comparing two directions of the same kind throughout.

How do I know whether to expect a single intersection point, no intersection, or infinitely many?

Substitute the line's parametrised point into the plane's equation and collect terms in $\lambda$ . A nonzero coefficient of $\lambda$ always gives exactly one solvable value of $\lambda$ , hence one intersection point. If the coefficient is zero, look at the constant left over: zero means the whole line satisfies the plane's equation (infinitely many points, the line lies in the plane), and nonzero means no value of $\lambda$ works (the line is parallel to the plane with no intersection).

Do I always need the absolute value in the sine formula, even if the numbers work out positive anyway?

Yes, write it in every time. The angle $\phi$ between a line and a plane is conventionally reported as the acute angle, so $\sin\phi$ must stay between $0$ and $1$ . Depending on how you wrote the line's direction ratios, the raw dot product $aa'+bb'+cc'$ can come out negative even though the geometric angle is unchanged, and dropping the absolute value would give you a negative sine, which is meaningless for an acute angle.

Angles and intersections

The core idea

Worked example

Competitive stretch

Common traps

Before this