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

Question

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

Solution — Step by Step

The line has direction ratios $\vec{b} = (2, 3, 6)$ .

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

The angle $\phi$ between a line and a plane is related to the angle $\theta$ between the line and the normal by: $\phi = 90° - \theta$ .

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

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

|\vec{b}| = \sqrt{4 + 9 + 36} = 7, \quad |\vec{n}| = \sqrt{9 + 16 + 144} = 13

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

\phi = \sin^{-1}\left(\frac{90}{91}\right) \approx \mathbf{81.9°}

Why This Works

graph TD
    A["3D Geometry Problem Type"] --> B["Angle between two lines"]
    A --> C["Angle between line and plane"]
    A --> D["Angle between two planes"]
    A --> E["Distance from point to plane"]
    B --> F["cos θ = |b₁·b₂| / |b₁||b₂|"]
    C --> G["sin φ = |b·n| / |b||n|"]
    D --> H["cos θ = |n₁·n₂| / |n₁||n₂|"]
    E --> I["d = |ax₁+by₁+cz₁-d| / √a²+b²+c²"]

The key insight: a line makes angle $\phi$ with a plane if and only if it makes angle $(90° - \phi)$ with the normal to the plane. So $\sin\phi = \cos\theta$ , which equals the dot product formula.

Lines in 3D can be written in three forms:

Cartesian: $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$
Vector: $\vec{r} = \vec{a} + \lambda\vec{b}$
Parametric: $x = x_1 + at$ , $y = y_1 + bt$ , $z = z_1 + ct$

Planes can be written as: $ax + by + cz = d$ or $\vec{r} \cdot \vec{n} = d$ .

Alternative Method

For JEE problems asking whether a line is parallel to a plane: check if $\vec{b} \cdot \vec{n} = 0$ . If the dot product is zero, the line is parallel to the plane (perpendicular to the normal).

For a line to LIE IN a plane, two conditions: $\vec{b} \cdot \vec{n} = 0$ AND the given point on the line must satisfy the plane equation. Many JEE problems test exactly this distinction.

Common Mistake

Using $\cos\phi$ instead of $\sin\phi$ for line-plane angle. The angle between a line and a plane uses $\sin\phi$ , not $\cos\phi$ . The angle between two planes or two lines uses $\cos\theta$ . This is because the line-plane angle is measured from the plane surface, not from the normal. Mixing up sin and cos here gives the complementary angle — and costs you the mark.

Angle between lines: $\cos\theta = \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$

Angle between line and plane: $\sin\phi = \frac{|al + bm + cn|}{\sqrt{a^2+b^2+c^2}\sqrt{l^2+m^2+n^2}}$

Distance from point to plane: $d = \frac{|ax_1 + by_1 + cz_1 - d|}{\sqrt{a^2+b^2+c^2}}$

Shortest distance between skew lines: $d = \frac{|(\vec{a_2}-\vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}$

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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