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Chapter 11 · Three Dimensional Geometry · Go for it

Lines in space, the quick way

A line in space is a point plus a direction. Learn to read the direction, write the line, and use the angle and distance formulas. Ten kinds of question cover the chapter.

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The chapter in one minute

What the chapter is about, and what the exam asks.

This chapter is about straight lines in space. A line is fixed by one point on it and its direction.

The direction is given by three numbers: direction ratios a,b,ca, b, c⁠, or direction cosines l,m,nl, m, n⁠, which are the same ratios scaled so that l2+m2+n2=1l^2 + m^2 + n^2 = 1⁠.

The exam asks you to find direction cosines, write the equation of a line, find the angle between two lines, test if two lines are perpendicular or parallel, find a line perpendicular to two lines, and find the shortest distance between two lines.

Most questions are one of ten kinds, and each kind has a fixed recipe. You will learn all ten below. Nine are in the book. The tenth, the foot of a perpendicular, is not in this book, but CBSE sample papers ask it.

10question types
13formula cards
1check for direction cosines: their squares add up to 1
0shortest distance when two lines meet
Formulas

The formulas you must know

Learn these cards. Every question in the chapter is built from them.

Direction cosines
l=cos⁡α, m=cos⁡β,n=cos⁡γ\begin{aligned} l &= \cos\alpha,\ m = \cos\beta, \\ n &= \cos\gamma \end{aligned}

α\alpha⁠, β\beta⁠, γ\gamma are the angles the line makes with the positive directions of the xx⁠, yy and zz-axes.

They always fit together
l2+m2+n2=1l^2 + m^2 + n^2 = 1

Use it to check your answer, or to find a missing cosine (up to its sign).

From ratios to cosines
l=aa2+b2+c2l = \frac{a}{\sqrt{a^2 + b^2 + c^2}}

Same bottom for mm (top bb⁠) and nn (top cc⁠). You may change all three signs: that is the same line, pointing the other way.

Through two points
x2−x1, y2−y1,z2−z1\begin{aligned} &x_2 - x_1,\ y_2 - y_1, \\ &z_2 - z_1 \end{aligned}

These are direction ratios of the line through P(x1,y1,z1)P(x_1, y_1, z_1) and Q(x2,y2,z2)Q(x_2, y_2, z_2)⁠. Divide each by the length PQPQ to get the cosines.

Vector form of a line
r⃗=a⃗+λb⃗\vec r = \vec a + \lambda\vec b

a⃗\vec a is the position vector of a point on the line, b⃗\vec b is its direction, and λ\lambda is any real number.

Cartesian form of a line
x−x1a=y−y1b=z−z1c\begin{aligned} \frac{x - x_1}{a} &= \frac{y - y_1}{b} \\ &= \frac{z - z_1}{c} \end{aligned}

(x1,y1,z1)(x_1, y_1, z_1) is a point on the line, and a,b,ca, b, c are its direction ratios.

Angle between two lines
cos⁡θ=∣b⃗1⋅b⃗2∣∣b⃗1∣ ∣b⃗2∣\cos\theta = \frac{|\vec b_1\cdot\vec b_2|}{|\vec b_1|\,|\vec b_2|}

With direction ratios, the top is ∣a1a2+b1b2+c1c2∣|a_1a_2 + b_1b_2 + c_1c_2|⁠. The bars give the acute angle.

Perpendicular lines
a1a2+b1b2+c1c2=0a_1a_2 + b_1b_2 + c_1c_2 = 0

Multiply matching direction ratios and add. A total of 0 means a right angle.

Parallel lines
a1a2=b1b2=c1c2\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}

The direction ratios are in proportion.

Cross product
b⃗1×b⃗2=∣i^j^k^a1b1c1a2b2c2∣\begin{aligned} &\vec b_1\times\vec b_2 \\ &= \begin{vmatrix} \hat i & \hat j & \hat k \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} \end{aligned}

Expand along the top row. The middle term, with j^\hat j⁠, takes a minus sign. The answer is perpendicular to both lines.

Shortest distance, skew lines
d=∣(b⃗1×b⃗2)⋅(a⃗2−a⃗1)∣b⃗1×b⃗2∣∣d = \left|\frac{(\vec b_1\times\vec b_2)\cdot(\vec a_2 - \vec a_1)}{|\vec b_1\times\vec b_2|}\right|

For r⃗=a⃗1+λb⃗1\vec r = \vec a_1 + \lambda\vec b_1 and r⃗=a⃗2+μb⃗2\vec r = \vec a_2 + \mu\vec b_2⁠. If dd is 0, the lines meet.

Distance, parallel lines
d=∣b⃗×(a⃗2−a⃗1)∣∣b⃗∣d = \frac{|\vec b\times(\vec a_2 - \vec a_1)|}{|\vec b|}

For r⃗=a⃗1+λb⃗\vec r = \vec a_1 + \lambda\vec b and r⃗=a⃗2+μb⃗\vec r = \vec a_2 + \mu\vec b⁠, with the same b⃗\vec b⁠.

Any point of a line
(x1+aλ, y1+bλ,z1+cλ)\begin{aligned} (&x_1 + a\lambda,\ y_1 + b\lambda, \\ &z_1 + c\lambda) \end{aligned}

For the line through (x1,y1,z1)(x_1, y_1, z_1) with direction ratios a,b,ca, b, c⁠. Each number λ\lambda gives one point. Use it for the foot of a perpendicular, or to see where two lines meet.

Question types

The question types and how to crack each one

For each type: how to spot it, what to do, and one real NCERT question worked the exam way.

1. Direction cosines from angles or from ratios

Spot it: you are given the angles with the axes, or three direction ratios, and asked for the direction cosines.

  1. From angles: take the cosine of each angle. That is all.
  2. From ratios a,b,ca, b, c⁠: work out a2+b2+c2\sqrt{a^2 + b^2 + c^2}⁠.
  3. Divide each ratio by it, and simplify the fractions.

Exercise 11.1 Q3. If a line has the direction ratios −18,12,−4-18, 12, -4⁠, then what are its direction cosines?

Here a=−18a = -18⁠, b=12b = 12⁠, c=−4c = -4⁠.

a2+b2+c2=324+144+16=484=22\begin{aligned} &\sqrt{a^2 + b^2 + c^2} \\ &= \sqrt{324 + 144 + 16} \\ &= \sqrt{484} = 22 \end{aligned}

So l=−1822l = \frac{-18}{22}⁠, m=1222m = \frac{12}{22}⁠, n=−422n = \frac{-4}{22}⁠.

∴ −911,611,−211-\frac{9}{11}, \frac{6}{11}, -\frac{2}{11}

See a full exam answer · Go deeper: see it visually · Basics: direction cosines

In past papers (4): click to see
Board 2023 Q11 · Sample 2022-23 Q23(b) · Sample 2021-22 T2 Q4 · Sample 2020-21 Q13

2. The line through two points: subtract, then divide

Spot it: two points are given, and you need the direction of the line through them.

  1. Subtract the coordinates, second point minus first. These are direction ratios.
  2. For cosines, find the length PQPQ⁠: the square root of the sum of their squares.
  3. Divide each ratio by PQPQ⁠.

Example 3. Find the direction cosines of the line passing through the two points (−2,4,−5)(-2, 4, -5) and (1,2,3)(1, 2, 3)⁠.

Direction ratios: 1−(−2)=31 - (-2) = 3⁠, 2−4=−22 - 4 = -2⁠, 3−(−5)=83 - (-5) = 8⁠.

PQ=32+(−2)2+82=9+4+64=77\begin{aligned} PQ &= \sqrt{3^2 + (-2)^2 + 8^2} \\ &= \sqrt{9 + 4 + 64} = \sqrt{77} \end{aligned}

∴ 377,−277,877\frac{3}{\sqrt{77}}, \frac{-2}{\sqrt{77}}, \frac{8}{\sqrt{77}}

See a full exam answer · Go deeper: see it visually · Basics: distance in 3D

In past papers (8): click to see
Board 2025 Q34(a) · Board 2025 Q34(b) · Board 2023 Q20 · Board 2023 Q34(b) · Sample 2026-27 Q23(a) · Sample 2026-27 Q30(a) · Sample 2024-25 Q35(b) · Sample 2022-23 Q18

3. Collinear points: two pieces in the same direction

Spot it: "Show that the points are collinear", that is, they lie on one straight line.

  1. Find the direction ratios of AB and of BC.
  2. Show they are in proportion: divide matching numbers and get the same answer each time.
  3. Conclude in words: AB is parallel to BC and B is on both, so the points are collinear.

Exercise 11.1 Q4. Show that the points (2,3,4)(2, 3, 4)⁠, (−1,−2,1)(-1, -2, 1)⁠, (5,8,7)(5, 8, 7) are collinear.

Call them A, B and C. Direction ratios of AB: −1−2,−2−3,1−4-1 - 2, -2 - 3, 1 - 4⁠, that is −3,−5,−3-3, -5, -3⁠. Of BC: 5+1,8+2,7−15 + 1, 8 + 2, 7 - 1⁠, that is 6,10,66, 10, 6⁠.

6−3=10−5=6−3=−2\frac{6}{-3} = \frac{10}{-5} = \frac{6}{-3} = -2

So AB is parallel to BC, and B is on both.

∴ A,B,CA, B, C are collinear.

See a full exam answer · Book Example 5 · Go deeper: see it visually · Basics: points in space

In past papers (1): click to see
Board 2026 Q14

4. Write the line: a point and a direction

Spot it: "Find the equation of the line through the point … parallel to (or in the direction of) …".

  1. Write a⃗\vec a for the point and b⃗\vec b for the direction.
  2. Vector form: r⃗=a⃗+λb⃗\vec r = \vec a + \lambda\vec b⁠.
  3. Cartesian form: subtract the point's coordinates on top, put the direction ratios at the bottom.

Exercise 11.2 Q5. Find the equation of the line in vector and in cartesian form that passes through the point with position vector 2i^−j^+4k^2\hat i - \hat j + 4\hat k and is in the direction i^+2j^−k^\hat i + 2\hat j - \hat k⁠.

Here a⃗=2i^−j^+4k^\vec a = 2\hat i - \hat j + 4\hat k and b⃗=i^+2j^−k^\vec b = \hat i + 2\hat j - \hat k⁠. The point is (2,−1,4)(2, -1, 4) and the direction ratios are 1,2,−11, 2, -1⁠, so the Cartesian form is

x−21=y+12=z−4−1\begin{aligned} \frac{x - 2}{1} &= \frac{y + 1}{2} \\ &= \frac{z - 4}{-1} \end{aligned}

The vector form is r⃗=a⃗+λb⃗\vec r = \vec a + \lambda\vec b⁠, where λ\lambda is any real number:

∴ r⃗=2i^−j^+4k^+λ(i^+2j^−k^)\vec r = 2\hat i - \hat j + 4\hat k + \lambda(\hat i + 2\hat j - \hat k)

See a full exam answer · Book Example 6 · Go deeper: see it visually · Basics: the vector equation of a line

In past papers (2): click to see
Board 2023 Q34(a) · Sample 2025-26 Q29

5. From Cartesian to vector form: read off the point and the direction

Spot it: a line is given as three equal fractions, and you need its vector form, its point or its direction.

  1. Make sure each top starts with xx⁠, yy or zz alone, with nothing in front.
  2. The point: the numbers on top, with their signs changed.
  3. The direction ratios: the numbers at the bottom. Then write r⃗=a⃗+λb⃗\vec r = \vec a + \lambda\vec b⁠.

Exercise 11.2 Q7. The cartesian equation of a line is x−53=y+47=z−62\frac{x - 5}{3} = \frac{y + 4}{7} = \frac{z - 6}{2}⁠. Write its vector form.

The point is (5,−4,6)(5, -4, 6) and the direction ratios are 3,7,23, 7, 2⁠. So

a⃗=5i^−4j^+6k^b⃗=3i^+7j^+2k^\begin{aligned} \vec a &= 5\hat i - 4\hat j + 6\hat k \\ \vec b &= 3\hat i + 7\hat j + 2\hat k \end{aligned}

∴ r⃗=5i^−4j^+6k^+λ(3i^+7j^+2k^)\vec r = 5\hat i - 4\hat j + 6\hat k + \lambda(3\hat i + 7\hat j + 2\hat k)

See a full exam answer · Go deeper: see it visually · Basics: a line is a point plus a direction

In past papers (5): click to see
Board 2023 Q11 · Sample 2026-27 Q30(a) · Sample 2025-26 Q35 · Sample 2022-23 Q23(b) · Sample 2021-22 T2 Q4

6. The angle between two lines: only the directions matter

Spot it: "Find the angle between the lines". The points on the lines play no part.

  1. Pick out the two directions: b⃗1\vec b_1 and b⃗2\vec b_2⁠, or the bottoms of the two Cartesian forms.
  2. Top: multiply matching numbers, add, and take the modulus.
  3. Bottom: the two lengths. The angle is cos⁡−1\cos^{-1} of the answer.

Exercise 11.2 Q9 (ii). Find the angle between the lines x2=y2=z1\frac{x}{2} = \frac{y}{2} = \frac{z}{1} and x−54=y−21=z−38\frac{x - 5}{4} = \frac{y - 2}{1} = \frac{z - 3}{8}⁠.

The direction ratios are 2,2,12, 2, 1 and 4,1,84, 1, 8⁠.

∣b⃗1∣=4+4+1=3∣b⃗2∣=16+1+64=9\begin{aligned} |\vec b_1| &= \sqrt{4 + 4 + 1} = 3 \\ |\vec b_2| &= \sqrt{16 + 1 + 64} = 9 \end{aligned} cos⁡θ=∣2(4)+2(1)+1(8)∣3×9=1827=23\begin{aligned} \cos\theta &= \frac{|2(4) + 2(1) + 1(8)|}{3\times 9} \\ &= \frac{18}{27} = \frac{2}{3} \end{aligned}

∴ θ=cos⁡−123\theta = \cos^{-1}\frac{2}{3}

See a full exam answer · Book Examples 7 and 8 · Go deeper: see it visually · Basics: the dot product

In past papers (2): click to see
Sample 2026-27 Q30(b) · Sample 2022-23 Q20

7. Perpendicular or parallel: test the direction ratios

Spot it: "Show that the lines are perpendicular (or parallel)", or "find kk so that the lines are at right angles".

  1. Write down the two sets of direction ratios.
  2. Perpendicular: put a1a2+b1b2+c1c2a_1a_2 + b_1b_2 + c_1c_2 equal to 0, and solve for the unknown.
  3. Parallel: the ratios must be in proportion. Show it, or use it to find an unknown.
  4. For a "show that", end with the conclusion in words.

Miscellaneous Q3. If the lines x−1−3=y−22k=z−32\frac{x - 1}{-3} = \frac{y - 2}{2k} = \frac{z - 3}{2} and x−13k=y−11=z−6−5\frac{x - 1}{3k} = \frac{y - 1}{1} = \frac{z - 6}{-5} are perpendicular, find the value of kk⁠.

The direction ratios are −3,2k,2-3, 2k, 2 and 3k,1,−53k, 1, -5⁠. The lines are perpendicular, so

−3(3k)+2k(1)+2(−5)=0−9k+2k−10=0−7k=10\begin{aligned} -3(3k) + 2k(1) + 2(-5) &= 0 \\ -9k + 2k - 10 &= 0 \\ -7k &= 10 \end{aligned}

∴ k=−107k = -\frac{10}{7}

See a full exam answer · Go deeper: see it visually · Basics: the dot product

In past papers (7): click to see
Board 2026 Q20 · Board 2026 Q35 · Board 2023 Q23(b) · Sample 2026-27 Q23(a) · Sample 2026-27 Q36(i) · Sample 2026-27 Q36(ii) · Sample 2023-24 Q5

8. The shortest distance between two lines

Spot it: "Find the shortest distance between the lines". First look at the two directions.

  1. Write a⃗1\vec a_1⁠, b⃗1\vec b_1⁠, a⃗2\vec a_2⁠, b⃗2\vec b_2⁠. If the lines are in Cartesian form, read them off first.
  2. Directions not in proportion (not parallel): find a⃗2−a⃗1\vec a_2 - \vec a_1 and b⃗1×b⃗2\vec b_1\times\vec b_2⁠, then use the skew formula.
  3. Same direction (parallel lines): use the parallel formula with b⃗×(a⃗2−a⃗1)\vec b\times(\vec a_2 - \vec a_1)⁠.
  4. Take the modulus. A distance is never negative.

Miscellaneous Q4. Find the shortest distance between the lines r⃗=6i^+2j^+2k^+λ(i^−2j^+2k^)\vec r = 6\hat i + 2\hat j + 2\hat k + \lambda(\hat i - 2\hat j + 2\hat k) and r⃗=−4i^−k^+μ(3i^−2j^−2k^)\vec r = -4\hat i - \hat k + \mu(3\hat i - 2\hat j - 2\hat k)⁠.

a⃗2−a⃗1=−10i^−2j^−3k^\begin{aligned} \vec a_2 - \vec a_1 &= -10\hat i - 2\hat j - 3\hat k \end{aligned} b⃗1×b⃗2=∣i^j^k^1−223−2−2∣=8i^+8j^+4k^\begin{aligned} \vec b_1\times\vec b_2 &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & -2 & 2 \\ 3 & -2 & -2 \end{vmatrix} \\ &= 8\hat i + 8\hat j + 4\hat k \end{aligned} ∣b⃗1×b⃗2∣=64+64+16=12\begin{aligned} |\vec b_1\times\vec b_2| &= \sqrt{64 + 64 + 16} \\ &= 12 \end{aligned} d=∣(b⃗1×b⃗2)⋅(a⃗2−a⃗1)∣b⃗1×b⃗2∣∣=∣−80−16−1212∣=10812\begin{aligned} d &= \left|\frac{(\vec b_1\times\vec b_2)\cdot(\vec a_2 - \vec a_1)}{|\vec b_1\times\vec b_2|}\right| \\ &= \left|\frac{-80 - 16 - 12}{12}\right| = \frac{108}{12} \end{aligned}

∴ d=9d = 9

See a full exam answer · Book Examples 9 and 10 · Go deeper: see it visually · Basics: the cross product

In past papers (9): click to see
Board 2025 Q29(a) · Board 2023 Q34(a) · Sample 2026-27 Q36(iii)(a) · Sample 2026-27 Q36(iii)(b) · Sample 2024-25 Q35(a) · Sample 2023-24 Q35(b) · Sample 2022-23 Q34 · Sample 2021-22 T2 Q10(a) · Sample 2020-21 Q37(a)

9. A line perpendicular to two lines: use the cross product

Spot it: "Find the line through the point … perpendicular to the two lines …". The CBSE sample papers for 2024-25, 2025-26 and 2026-27 all ask it.

  1. Read the two directions b⃗1\vec b_1 and b⃗2\vec b_2⁠.
  2. Work out b⃗1×b⃗2\vec b_1\times\vec b_2⁠. It is perpendicular to both, so it is the new direction. Divide out a common factor.
  3. Write the line through the given point with this direction, in the form asked.

CBSE Sample Paper 2024-25, Q28 (b). Find the vector and the cartesian equation of the line that passes through (−1,2,7)(-1, 2, 7) and is perpendicular to the lines r⃗=2i^+j^−3k^+λ(i^+2j^+5k^)\vec r = 2\hat i + \hat j - 3\hat k + \lambda(\hat i + 2\hat j + 5\hat k) and r⃗=3i^+3j^−7k^+μ(3i^−2j^+5k^)\vec r = 3\hat i + 3\hat j - 7\hat k + \mu(3\hat i - 2\hat j + 5\hat k)⁠.

The directions are b⃗1=i^+2j^+5k^\vec b_1 = \hat i + 2\hat j + 5\hat k and b⃗2=3i^−2j^+5k^\vec b_2 = 3\hat i - 2\hat j + 5\hat k⁠. The new line is perpendicular to both, so its direction is

b⃗1×b⃗2=∣i^j^k^1253−25∣=20i^+10j^−8k^\begin{aligned} \vec b_1\times\vec b_2 &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & 2 & 5 \\ 3 & -2 & 5 \end{vmatrix} \\ &= 20\hat i + 10\hat j - 8\hat k \end{aligned}

Take 10,5,−410, 5, -4 as its direction ratios. The Cartesian form is

x+110=y−25=z−7−4\begin{aligned} \frac{x + 1}{10} &= \frac{y - 2}{5} \\ &= \frac{z - 7}{-4} \end{aligned}

The vector form is

∴ r⃗=−i^+2j^+7k^+λ(10i^+5j^−4k^)\vec r = -\hat i + 2\hat j + 7\hat k + \lambda(10\hat i + 5\hat j - 4\hat k)

See a similar one worked step by step (Miscellaneous Q5) · Go deeper: see it visually · Basics: the cross product

In past papers (3): click to see
Sample 2026-27 Q30(b) · Sample 2025-26 Q35 · Sample 2024-25 Q28(b)

10. Beyond the book, but in CBSE papers: the foot of a perpendicular

Spot it: "Find the foot of the perpendicular from a point to a line", "the image of a point in a line", or "do the lines meet?". This book has none of these, but recent CBSE sample papers ask all three.

  1. Write any point Q of the line with λ\lambda⁠: (x1+aλ,y1+bλ,z1+cλ)(x_1 + a\lambda, y_1 + b\lambda, z_1 + c\lambda)⁠.
  2. Foot from P: PQ must be perpendicular to the line. Multiply the direction ratios of PQ and of the line, add, put the total equal to 0, and solve for λ\lambda⁠. That Q is the foot, and PQ is the distance.
  3. Image of P: the foot is the midpoint of P and its image.
  4. Do two lines meet? Write any point of each (with λ\lambda and μ\mu⁠), set them equal, solve two of the equations, and check the third.

Beyond the book B3. Find the foot of the perpendicular from P(1,6,3)P(1, 6, 3) to the line x1=y−12=z−23\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}⁠, and the distance of P from the line.

Any point of the line is Q(λ,2λ+1,3λ+2)Q(\lambda, 2\lambda + 1, 3\lambda + 2)⁠. PQ has direction ratios λ−1,2λ−5,3λ−1\lambda - 1, 2\lambda - 5, 3\lambda - 1⁠, and the line has 1,2,31, 2, 3⁠. They are perpendicular, so

(λ−1)+2(2λ−5)+3(3λ−1)=0\begin{aligned} &(\lambda - 1) + 2(2\lambda - 5) \\ &\quad + 3(3\lambda - 1) = 0 \end{aligned} 14λ−14=0λ=1\begin{aligned} 14\lambda - 14 &= 0 \\ \lambda &= 1 \end{aligned}

So the foot is Q(1,3,5)Q(1, 3, 5)⁠, and PQ has direction ratios 0,−3,20, -3, 2⁠.

PQ=0+9+4=13\begin{aligned} PQ &= \sqrt{0 + 9 + 4} \\ &= \sqrt{13} \end{aligned}

∴ foot (1,3,5)(1, 3, 5)⁠, distance 13\sqrt{13} units

See it worked step by step · Go deeper: see it visually · Basics: the dot product

In past papers (12): click to see
Board 2026 Q16 · Board 2026 Q35 · Board 2025 Q34(a) · Board 2024 Q16 · Board 2024 Q35 · Board 2022 T2 Q14(ii) · Sample 2026-27 Q30(a) · Sample 2025-26 Q14 · Sample 2025-26 Q29 · Sample 2024-25 Q35(b) · Sample 2023-24 Q35(a) · Sample 2020-21 Q37(a)
Shortcuts

Shortcuts and quick tips

Small habits that save time and marks.

  • Direction ratios can be scaled. 2,−4,62, -4, 6 and 1,−2,31, -2, 3 give the same direction, so divide out a common factor first to keep the numbers small.
  • Check direction cosines by squaring and adding: 81121+36121+4121=1\frac{81}{121} + \frac{36}{121} + \frac{4}{121} = 1⁠. If you do not get 1, look for a slip.
  • Equal angles with all three axes: l=m=nl = m = n⁠, so 3l2=13l^2 = 1 and l=m=n=±13l = m = n = \pm\frac{1}{\sqrt 3}⁠.
  • Before you read a Cartesian line, make each top start with xx⁠, yy or zz alone: 1−x3=x−1−3\frac{1 - x}{3} = \frac{x - 1}{-3} and 7y−142p=y−22p/7\frac{7y - 14}{2p} = \frac{y - 2}{2p/7}⁠.
  • The angle needs only the directions. Ignore the points on the lines.
  • If the top of the angle formula is 0, stop: the lines are perpendicular and the angle is 90∘90^\circ⁠.
  • Look at the two directions before a distance question. The same b⃗\vec b (or one a multiple of the other) means parallel lines: use the parallel formula, because b⃗1×b⃗2\vec b_1\times\vec b_2 is then 0⃗\vec 0⁠.
  • Skew lines are lines that are neither parallel nor meeting. If the directions are not in proportion and the shortest distance is not 0, the lines are skew.
  • A line along an axis has zeros in its direction ratios: the xx-axis has 1,0,01, 0, 0⁠. A 0 at the bottom of a Cartesian form just means that coordinate never changes, as in x1=y0=z0\frac{x}{1} = \frac{y}{0} = \frac{z}{0}⁠.
  • No calculator: leave answers such as cos⁡−123\cos^{-1}\frac{2}{3} and 2937\frac{\sqrt{293}}{7} exactly as they are.
Watch out

Mistakes that cost marks

Each one is common, and each has an easy fix.

  • Reading the point of x+23\frac{x + 2}{3} as 2 → change the sign: it is −2-2⁠.
  • Taking 3 as the direction ratio in 1−x3\frac{1 - x}{3} → rewrite it as x−1−3\frac{x - 1}{-3}⁠: the ratio is −3-3⁠.
  • Giving the direction ratios when cosines are asked for → divide each one by a2+b2+c2\sqrt{a^2 + b^2 + c^2}⁠.
  • Dropping the minus on the middle term of a cross product → expand as i^(…)−j^(…)+k^(…)\hat i(\ldots) - \hat j(\ldots) + \hat k(\ldots)⁠.
  • Subtracting the directions instead of the points → the distance formulas use a⃗2−a⃗1\vec a_2 - \vec a_1⁠, the two points.
  • A negative answer for cos⁡θ\cos\theta or for a distance → keep the modulus bars; with them, neither can be negative.
  • Using the skew formula for parallel lines → it divides by 0; use the parallel formula instead.
Marks

How the examiner gives marks

From CBSE's own marking schemes and curriculum. Write these lines and the marks follow.

  • Vectors and three dimensional geometry together carry 14 of the 80 marks in the theory paper.
  • Marks are given step by step for value points. Write a⃗1\vec a_1⁠, b⃗1\vec b_1⁠, a⃗2\vec a_2⁠, b⃗2\vec b_2 (or the direction ratios), the formula, the working and the answer, each on its own line.
  • In a "show that" question, end with the conclusion in words, such as "so the lines are perpendicular". CBSE gives marks for conclusions in words.
  • Give the final answer exactly, with no decimals: θ=cos⁡−123\theta = \cos^{-1}\frac{2}{3}⁠, or d=2937d = \frac{\sqrt{293}}{7} units.
  • A slip is penalised only once, so keep going. Any correct method gets full marks.
  • An MCQ or assertion-reason question earns 1 mark for the right option only.
Practice

Practise these first

Easiest first. Together they cover every type above.

  1. Exercise 11.1 Q2 equal angles with the axes
  2. Exercise 11.1 Q5 the three sides of a triangle: subtract, then divide (the key prints 317\frac{3}{17} for 317\frac{3}{\sqrt{17}}⁠)
  3. Example 5 three collinear points
  4. Exercise 11.2 Q6 read the direction of one line, write the Cartesian form of another
  5. Exercise 11.2 Q8 the angle between two lines in vector form
  6. Exercise 11.2 Q11 show that two lines are perpendicular
  7. Exercise 11.2 Q3 show that two lines are parallel
  8. Exercise 11.2 Q10 find pp⁠: rewrite the tops first
  9. Miscellaneous Q5 a line perpendicular to two lines
  10. Exercise 11.2 Q14 the shortest distance between skew lines
  11. Example 10 the distance between parallel lines
  12. Beyond the book B2 do two lines meet, and where
Next

You can do this

Ten kinds of question, one recipe each. Read the direction, write the line, and use the formula cards. Work through the practice list and you are ready for the exam.

Ready to go deeper? Open the interactive lesson · Four more problems beyond the book, of the kinds CBSE papers ask

Past papers

35 CBSE board and sample-paper questions on Three Dimensional Geometry, each with a full-marks exam answer. Practise them →