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JEE Mains Maths · Three Dimensional Geometry

Lines Meeting Planes

A line and a plane together: where the line meets the plane and in what ratio a plane cuts a segment, the angle between them, when a line is parallel to or lies in a plane, and distances measured along a given direction.

All Three Dimensional Geometry notesJEE Mains Maths strategy

Why this matters

Thirty-six PYQs. Almost every one substitutes the line's general point into the plane's equation. The largest group measures a distance 'parallel to a line', which is the same substitution read as a length. Three ideas cover the page.

Concept 1 of 3: Where a line meets a plane, and the ratio a plane cuts

Put the line's general point A+td⃗A+t\vec d into the plane's equation. It becomes one linear equation in tt, and its root gives the meeting point. The same idea tells how a plane cuts a segment ABAB: the plane's values at AA and BB are in the ratio of the pieces, and opposite signs mean the cut is between them.

Definition

  • Meeting point: substitute A+td⃗A+t\vec d into the plane and solve for tt.
  • Ratio: a plane divides ABAB in the ratio −P(A)P(B)-\frac{P(A)}{P(B)}, internally when P(A)P(A), P(B)P(B) have opposite signs.
  • Projection of a line on a plane: it passes through the meeting point and through the foot of any other point of the line.

Ratio in which a plane divides AB

ACCB=−P(A)P(B)\frac{AC}{CB}=-\frac{P(A)}{P(B)}

Worked example

Where does x−11=y−22=z−33\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3} meet x+y+z=12x+y+z=12?
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 22 · Q83Moderate

Example 1 · Three Dimensional Geometry · Lines Meeting Planes

The square of the distance of the point of intersection of the line x−12=y−23=z+16\frac{x- 1}{2}=\frac{y- 2}{3}=\frac{z+ 1}{6} and the plane 2x−y+z=62x - y + z = 6 from the point (−1,−1,2)( - 1, - 1,2) is

0⋅t=c0\cdot t=c means no meeting point

If tt drops out, the line is parallel to the plane: no solution when the constant is non-zero, every point when it is zero (the line lies in the plane).
Drill 10 more on where a line meets a plane, and the ratio a plane cuts

Concept 2 of 3: The angle between a line and a plane; parallel and lying in

The angle a line makes with a plane is the complement of the angle it makes with the normal, so it uses SINE: sin⁡θ=∣d⃗⋅n⃗∣∣d⃗∣∣n⃗∣\sin\theta=\frac{|\vec d\cdot\vec n|}{|\vec d||\vec n|}. A line is parallel to the plane when d⃗⊥n⃗\vec d\perp\vec n; it lies in the plane when, in addition, one of its points does. A line given as two planes has direction n⃗1×n⃗2\vec n_1\times\vec n_2.

Definition

  • sin⁡θ=∣d⃗⋅n⃗∣∣d⃗∣∣n⃗∣\sin\theta=\frac{|\vec d\cdot\vec n|}{|\vec d||\vec n|}.
  • Parallel: d⃗⋅n⃗=0\vec d\cdot\vec n=0. Lies in the plane: also one point satisfies it.
  • Line of intersection of two planes: direction n⃗1×n⃗2\vec n_1\times\vec n_2.
  • A line parallel to a plane is at a constant distance from it: the distance of any of its points.

Angle between a line and a plane

sin⁡θ=∣d⃗⋅n⃗∣∣d⃗∣ ∣n⃗∣\sin\theta=\frac{|\vec d\cdot\vec n|}{|\vec d|\,|\vec n|}

Worked example

Find the angle between x1=y2=z2\frac x1=\frac y2=\frac z2 and the plane 2x−y+2z=52x-y+2z=5.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 11 April 2023 · Q178Moderate

Example 2 · Three Dimensional Geometry · Lines Meeting Planes

Let the line l:x=1−y−2=z−3λ,λ∈R\mathcal{l:}x = \frac{1 - y}{- 2} = \frac{z - 3}{\lambda},\lambda \in R meet the plane P:x+2y+3z=4P:x + 2y + 3z = 4 at the point (α,β,γ)(\alpha,\beta,\gamma). If the angle between the line l\mathcal{l} and the plane PP is cos⁡−1(514)\cos^{- 1}\left( \sqrt{\frac{5}{14}} \right), then α+2β+6γ\alpha + 2\beta + 6\gamma is equal to

SINE for a line and a plane

The dot product with the NORMAL gives the angle with the normal. The angle with the plane is its complement, so the formula has sin⁡θ\sin\theta. A question stating cos⁡θ\cos\theta of the line-plane angle needs converting first.
Drill 9 more on the angle between a line and a plane; parallel and lying in

Concept 3 of 3: Distance measured parallel to a given line

'The distance of PP from a plane measured parallel to a line' means: leave PP in the given direction until you reach the plane, and measure that walk. Write P+td⃗P+t\vec d, substitute into the plane to find tt, and the distance is ∣t∣ ∣d⃗∣|t|\,|\vec d|. If the target is a line instead of a plane, make the point P+td⃗P+t\vec d satisfy that line's equations.

Definition

  • Moving point: P+td⃗P+t\vec d, d⃗\vec d the given direction.
  • Target a plane: substitute and solve for tt.
  • Target a line: equate with the line's general point; two coordinates give the parameters, the third checks.
  • Distance: ∣t∣ ∣d⃗∣|t|\,|\vec d| (or ∣t∣|t| if d⃗\vec d was a unit vector).
  • If d⃗\vec d is parallel to the plane, there is no such distance.

Distance along a direction

distance=∣t∣ ∣d⃗∣,P+td⃗ on the target\text{distance}=|t|\,|\vec d|,\quad P+t\vec d\ \text{on the target}

Worked example

Find the distance of (1,2,3)(1,2,3) from x+y+z=12x+y+z=12 measured parallel to (2,2,1)(2,2,1).
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 20 · Q66Moderate

Example 3 · Three Dimensional Geometry · Lines Meeting Planes

The distance of the point (1,−2,3)(1, - 2,3) from the plane x−y+z=5x - y + z = 5 measured parallel to a line, whose direction ratios are 2,3,−62,3, - 6 is:

Multiply by ∣d⃗∣|\vec d| at the end

tt counts steps of d⃗\vec d, not units of length. Unless d⃗\vec d is a unit vector, the distance is ∣t∣ ∣d⃗∣|t|\,|\vec d|.
Drill 14 more on distance measured parallel to a given line

Summary — formulas & gotchas at a glance

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Formulas (3)

Watch out for (3)

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