3D geometry — line equations, plane equations, angle between them

Question

Find the angle between the line $\dfrac{x - 1}{2} = \dfrac{y + 1}{3} = \dfrac{z - 2}{6}$ and the plane $3x + 4y + 12z = 7$ .

(CBSE 12 Board / JEE Main standard problem)

Problem Type Decision Tree

flowchart TD
    A["3D Geometry Problem"] --> B{What is given?}
    B -->|Two lines| C{Find what?}
    B -->|Two planes| D{Find what?}
    B -->|Line + Plane| E{Find what?}
    C -->|Angle| C1["cos theta = |b1.b2| / |b1||b2|"]
    C -->|Shortest distance| C2["Use cross product formula"]
    D -->|Angle| D1["cos theta = |n1.n2| / |n1||n2|"]
    D -->|Distance| D2["Point-to-plane formula"]
    E -->|Angle| E1["sin theta = |b.n| / |b||n|"]
    E -->|Point of intersection| E2["Parametric form, substitute"]

Solution — Step by Step

The line has direction ratios $\vec{b} = (2, 3, 6)$ (the denominators in symmetric form).

The plane $3x + 4y + 12z = 7$ has normal vector $\vec{n} = (3, 4, 12)$ .

The angle $\theta$ between a line and a plane is given by:

\sin\theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}||\vec{n}|}

Why $\sin$ and not $\cos$ ? Because $\theta$ is measured between the line and the plane (not the normal). The angle between the line and the normal is $90° - \theta$ , so what would be $\cos$ with the normal becomes $\sin$ with the plane.

\vec{b} \cdot \vec{n} = 2(3) + 3(4) + 6(12) = 6 + 12 + 72 = 90

|\vec{b}| = \sqrt{4 + 9 + 36} = \sqrt{49} = 7

|\vec{n}| = \sqrt{9 + 16 + 144} = \sqrt{169} = 13

\sin\theta = \frac{90}{7 \times 13} = \frac{90}{91}

\boxed{\theta = \sin^{-1}\left(\frac{90}{91}\right)}

Why This Works

A plane is defined by its normal direction. The angle between a line and a plane is complementary to the angle between the line and the normal. Since the dot product naturally gives the cosine of the angle between two vectors, and we need the complement, we use sine.

If $\phi$ is the angle between $\vec{b}$ and $\vec{n}$ , then $\cos\phi = \vec{b}\cdot\vec{n}/(|\vec{b}||\vec{n}|)$ , and $\theta = 90° - \phi$ , so $\sin\theta = \cos\phi$ .

Alternative Method — Using the Normal Angle

Find the angle between the line and the normal first:

\cos\phi = \frac{90}{91}

Then $\theta = 90° - \phi = 90° - \cos^{-1}(90/91)$ .

This is algebraically identical but some students find it more intuitive to work with the normal first and then subtract from $90°$ .

In JEE Main, 3D geometry questions almost always test one of three things: (1) angle between line and plane, (2) shortest distance between skew lines, (3) image of a point in a plane. Memorise the formula for each — they appear as direct 2-mark questions.

Common Mistake

The classic error: using $\cos\theta$ instead of $\sin\theta$ for the line-plane angle. Students confuse the line-plane formula with the plane-plane formula. Remember: line-plane uses $\sin$ , plane-plane uses $\cos$ , line-line uses $\cos$ .

Another trap: forgetting the absolute value. The angle between a line and a plane is always between $0°$ and $90°$ , so we take the modulus of the dot product. Without it, you might report a negative angle, which is meaningless.