Three Dimensional Geometry Class 12: Notes, Methods and Solved Questions

What Three Dimensional Geometry is about

Weightage: This chapter is part of Unit IV, Vectors and Three-Dimensional Geometry, which carries 14 of the 80 theory marks together with Vector Algebra.Source: CBSE Mathematics syllabus 2026-27, Unit IV Vectors and Three-Dimensional Geometry

In real board papers: asked in 8 of the 8 board papers we hold (2023 to 2026), on average 8.3 marks per paper. See the questions

Key concepts

Direction cosines and direction ratios of a line

NCERT section 11.2

If a directed line makes angles α, β, γ with the positive x-, y- and z-axes, its direction cosines are $l=\cos \alpha$, $m=\cos \beta$, $n=\cos \gamma$, and always

$l^2+m^2+n^2=1$

Reversing the direction changes all three signs. Any numbers a, b, c proportional to l, m, n are direction ratios (infinitely many sets). From direction ratios:

$l=\pm \frac{a}{\sqrt{a^2+b^2+c^2}},m=\pm \frac{b}{\sqrt{a^2+b^2+c^2}},n=\pm \frac{c}{\sqrt{a^2+b^2+c^2}}$ (same sign throughout)

Worked example

A line makes angles 90°, 60° and 30° with the x-, y- and z-axes. Find its direction cosines. [1 mark]

Try it first, then show the solutionSolution

1. $l=\cos 90°=0$, $m=\cos 60°=\frac{1}{2}$, $n=\cos 30°=\frac{\sqrt{3}}{2}$
2. Check: $0+\frac{1}{4}+\frac{3}{4}=1$

Answer: $0,\frac{1}{2},\frac{\sqrt{3}}{2}$

Direction cosines of a line through two points

NCERT section 11.2.1

For the line through $P(x_1,y_1,z_1)$ and $Q(x_2,y_2,z_2)$, the direction ratios are

$x_2-x_1,y_2-y_1,z_2-z_1$

and dividing by the distance PQ gives the direction cosines. Two lines through pairs of points are parallel if their direction ratios are proportional, and perpendicular if $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$. Three points are collinear if the direction ratios of PQ and PR are proportional.

Worked example

Find the direction cosines of the line through P(−2, 4, −5) and Q(1, 2, 3). [2 marks]

Try it first, then show the solutionSolution

1. Direction ratios: 3, −2, 8; $PQ=\sqrt{9+4+64}=\sqrt{77}$

Answer: $\frac{3}{\sqrt{77}},-\frac{2}{\sqrt{77}},\frac{8}{\sqrt{77}}$

Vector equation of a line

NCERT section 11.3.1

A line through the point A with position vector $\vec{a}$, parallel to the vector $\vec{b}$, has vector equation

$\vec{r}=\vec{a}+\lambda \vec{b}$, λ a real parameter

Each value of λ gives one point of the line. $\vec{b}$ supplies the direction ratios; $\vec{a}$ is any one point on the line. The same line can be written with a different point or a multiple of $\vec{b}$.

Worked example

Write the vector equation of the line through (2, −1, 4) parallel to $\hat{i}+2\hat{j}-\hat{k}$. [1 mark]

Try it first, then show the solutionSolution

1. $\vec{a}=2\hat{i}-\hat{j}+4\hat{k}$, $\vec{b}=\hat{i}+2\hat{j}-\hat{k}$

Answer: $\vec{r}=(2\hat{i}-\hat{j}+4\hat{k})+\lambda (\hat{i}+2\hat{j}-\hat{k})$

Cartesian equation of a line and standard form

NCERT section 11.3.1

Through $(x_1,y_1,z_1)$ with direction ratios a, b, c:

$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$

In standard form the coefficients of x, y and z in the numerators are all +1. Before reading off the point and direction ratios, rewrite expressions like $\frac{4-y}{7}$ as $\frac{y-4}{-7}$ and $\frac{2z-6}{4}$ as $\frac{z-3}{2}$. To switch forms, put each fraction equal to λ: $x=x_1+a\lambda$, and so on.

Worked example

Write $\frac{x-5}{3}=\frac{4-y}{7}=\frac{2z-6}{4}$ in standard form, and give its vector form. [2 marks]

Try it first, then show the solutionSolution

1. $\frac{4-y}{7}=\frac{y-4}{-7}$; $\frac{2z-6}{4}=\frac{z-3}{2}$
2. Standard form: $\frac{x-5}{3}=\frac{y-4}{-7}=\frac{z-3}{2}$: point (5, 4, 3), direction ratios 3, −7, 2

Answer: $\vec{r}=(5\hat{i}+4\hat{j}+3\hat{k})+\lambda (3\hat{i}-7\hat{j}+2\hat{k})$

Line through two given points

NCERT section 11.3

A line through the points with position vectors $\vec{a}$ and $\vec{b}$ has direction $\vec{b}-\vec{a}$, so

$\vec{r}=\vec{a}+\lambda (\vec{b}-\vec{a})$, and in Cartesian form $\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$

This is just the point-and-direction form with direction ratios taken from the two points; simplify the ratios by a common factor if you like.

Worked example

Find the Cartesian equation of the line through A(2, −1, 1) and B(4, 3, −1). [1 mark]

Try it first, then show the solutionSolution

1. Direction ratios 2, 4, −2, or 1, 2, −1

Answer: $\frac{x-2}{1}=\frac{y+1}{2}=\frac{z-1}{-1}$

Angle between two lines

NCERT section 11.4

If two lines have direction ratios $a_1,b_1,c_1$ and $a_2,b_2,c_2$ (or directions $\overrightarrow{b_1}$, $\overrightarrow{b_2}$), the acute angle θ between them is given by

$\cos \theta =\frac{|a_1{a}_2+b_1{b}_2+c_1{c}_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}=\frac{|\overrightarrow{b_1}\cdot \overrightarrow{b_2}|}{|\overrightarrow{b_1}||\overrightarrow{b_2}|}$

• Perpendicular: $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$
• Parallel: $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

The points on the lines do not matter here, only the directions.

Worked example

Find the angle between two lines with direction ratios 2, 2, 1 and 4, 1, 8. [2 marks]

Try it first, then show the solutionSolution

1. $a_1{a}_2+b_1{b}_2+c_1{c}_2=8+2+8=18$; magnitudes 3 and 9
2. $\cos \theta =\frac{18}{27}=\frac{2}{3}$

Answer: $\theta ={\cos }^{-1}\frac{2}{3}$

General point, foot of a perpendicular and image of a point

NCERT section 11.3

Every point of $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$ can be written as $Q(x_1+a\lambda ,y_1+b\lambda ,z_1+c\lambda )$. To find the foot of the perpendicular from a point P to the line:

1. Write Q in terms of λ and form $\overrightarrow{PQ}$.
2. Use $\overrightarrow{PQ}\cdot \vec{b}=0$ (perpendicular to the line) to find λ, hence Q.
3. The perpendicular distance is PQ.
4. The image P′ of P in the line satisfies "Q is the mid-point of PP′", so $P^{\prime }=2Q-P$.

The same general-point idea finds a point on the line at a given distance from P (set $P{Q}^2$ equal to the square of the distance).

Worked example

Find the foot of the perpendicular from P(1, 3, 2) to the line $\frac{x-2}{1}=\frac{y+1}{2}=\frac{z-1}{2}$, the perpendicular distance and the image of P. [5 marks]

Try it first, then show the solutionSolution

1. General point $Q(2+\lambda ,-1+2\lambda ,1+2\lambda )$; $\overrightarrow{PQ}=(1+\lambda ,-4+2\lambda ,-1+2\lambda )$
2. $\overrightarrow{PQ}\cdot (1,2,2)=1+\lambda -8+4\lambda -2+4\lambda =9\lambda -9=0\Rightarrow \lambda =1$
3. Foot Q(3, 1, 3); $\overrightarrow{PQ}=(2,-2,1)$, so PQ = 3
4. Image $P^{\prime }=2Q-P=(5,-1,4)$

Answer: Foot (3, 1, 3); distance 3 units; image (5, −1, 4).

Skew lines and the shortest distance between them

NCERT section 11.5.1

Two lines in space can be parallel, intersecting, or skew: neither parallel nor intersecting (they lie in different planes, like a flyover and the road beneath it at an angle). The shortest distance is the length of the segment perpendicular to both lines.

For $\vec{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1}$ and $\vec{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}$:

$d=\left|\frac{(\overrightarrow{b_1}\times \overrightarrow{b_2})\cdot (\overrightarrow{a_2}-\overrightarrow{a_1})}{|\overrightarrow{b_1}\times \overrightarrow{b_2}|}\right|$

If d = 0 the lines intersect; if d > 0 (and the directions are not parallel) they are skew.

Worked example

Find the shortest distance between $\vec{r}=(\hat{i}+2\hat{k})+\lambda (\hat{i}+2\hat{j}+\hat{k})$ and $\vec{r}=(2\hat{i}+\hat{j}-\hat{k})+\mu (2\hat{i}-\hat{j}+\hat{k})$. [3 marks]

Try it first, then show the solutionSolution

1. $\overrightarrow{b_1}\times \overrightarrow{b_2}=\hat{i}(2+1)-\hat{j}(1-2)+\hat{k}(-1-4)=3\hat{i}+\hat{j}-5\hat{k}$, magnitude $\sqrt{35}$
2. $\overrightarrow{a_2}-\overrightarrow{a_1}=\hat{i}+\hat{j}-3\hat{k}$; dot product $=3+1+15=19$
3. $d=\frac{19}{\sqrt{35}}$

Answer: $\frac{19}{\sqrt{35}}$ units (the lines are skew)

Distance between parallel lines

NCERT section 11.5.2

If the lines are parallel, $\overrightarrow{b_1}\times \overrightarrow{b_2}=\vec{0}$ and the skew-line formula fails. For $\vec{r}=\overrightarrow{a_1}+\lambda \vec{b}$ and $\vec{r}=\overrightarrow{a_2}+\mu \vec{b}$:

$d=\frac{|\vec{b}\times (\overrightarrow{a_2}-\overrightarrow{a_1})|}{|\vec{b}|}$

If one direction is a multiple of the other (like $2\hat{i}-\hat{j}+2\hat{k}$ and $4\hat{i}-2\hat{j}+4\hat{k}$), use either as $\vec{b}$.

Worked example

Find the distance between $\vec{r}=(\hat{i}+2\hat{j}-\hat{k})+\lambda (2\hat{i}-\hat{j}+2\hat{k})$ and $\vec{r}=(3\hat{i}-\hat{j}+\hat{k})+\mu (4\hat{i}-2\hat{j}+4\hat{k})$. [3 marks]

Try it first, then show the solutionSolution

1. Parallel: take $\vec{b}=2\hat{i}-\hat{j}+2\hat{k}$, $|\vec{b}|=3$; $\overrightarrow{a_2}-\overrightarrow{a_1}=2\hat{i}-3\hat{j}+2\hat{k}$
2. $\vec{b}\times (\overrightarrow{a_2}-\overrightarrow{a_1})=\hat{i}(-2+6)-\hat{j}(4-4)+\hat{k}(-6+2)=4\hat{i}-4\hat{k}$, magnitude $4\sqrt{2}$
3. $d=\frac{4\sqrt{2}}{3}$

Answer: $\frac{4\sqrt{2}}{3}$ units

Do two lines intersect?

NCERT section 11.5

To decide and to find the point of intersection, equate general points: $\overrightarrow{a_1}+\lambda \overrightarrow{b_1}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}$ gives three equations in λ and μ. Solve two of them and test the third:

• third equation satisfied → the lines intersect; substitute λ to get the point;
• not satisfied (and directions not parallel) → the lines are skew.

Equivalently, the lines intersect exactly when the shortest distance is 0.

Worked example

Show that $\vec{r}=(\hat{i}-\hat{k})+\lambda (\hat{i}+\hat{j}+2\hat{k})$ and $\vec{r}=(4\hat{i}-\hat{j}+4\hat{k})+\mu (2\hat{i}-2\hat{j}+3\hat{k})$ intersect, and find the point. [3 marks]

Try it first, then show the solutionSolution

1. $1+\lambda =4+2\mu$, $\lambda =-1-2\mu$, $-1+2\lambda =4+3\mu$
2. From the first two: $1-1-2\mu =4+2\mu \Rightarrow \mu =-1$, λ = 1
3. Third: −1 + 2 = 1 and 4 − 3 = 1 ✓, so they intersect at λ = 1

Answer: They intersect at (2, 1, 1).

Maths in real life: Three Dimensional Geometry

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Equation of a Line in Space
• Foot of Perpendicular and Image of a Point
• Shortest Distance Between Two Lines

Three Dimensional Geometry formula sheet

Direction cosines

$l^2+m^2+n^2=1$

From direction ratios

$l=\frac{a}{\sqrt{a^2+b^2+c^2}},m=\frac{b}{\sqrt{a^2+b^2+c^2}},n=\frac{c}{\sqrt{a^2+b^2+c^2}}$

Direction ratios through two points

$x_2-x_1,y_2-y_1,z_2-z_1$

Vector equation of a line

$\vec{r}=\vec{a}+\lambda \vec{b}$

Cartesian equation of a line

$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$

Line through two points

$\vec{r}=\vec{a}+\lambda (\vec{b}-\vec{a}),\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$

Angle between lines

$\cos \theta =\frac{|a_1{a}_2+b_1{b}_2+c_1{c}_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$

Perpendicular and parallel lines

$a_1{a}_2+b_1{b}_2+c_1{c}_2=0;\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

Shortest distance (skew lines)

$d=\left|\frac{(\overrightarrow{b_1}\times \overrightarrow{b_2})\cdot (\overrightarrow{a_2}-\overrightarrow{a_1})}{|\overrightarrow{b_1}\times \overrightarrow{b_2}|}\right|$

Distance between parallel lines

$d=\frac{|\vec{b}\times (\overrightarrow{a_2}-\overrightarrow{a_1})|}{|\vec{b}|}$

Image of P in a line

$P^{\prime }=2Q-P,\text{where Q is the foot of the perpendicular}$

Distance of (a, b, c) from the axes

$\text{x-axis: }\sqrt{b^2+c^2},\text{y-axis: }\sqrt{a^2+c^2},\text{z-axis: }\sqrt{a^2+b^2}$

How the board asks Three Dimensional Geometry

The question patterns to expect, and the method that works for each.

Tricks to remember Three Dimensional Geometry

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Standard form before anything

  Make x, y, z appear with coefficient +1 in each numerator before reading the point or the direction ratios.

  Use it when: Every Cartesian line question.

• Squares add to one

  Direction cosines: l² + m² + n² = 1 always; direction ratios: no such rule.

  Use it when: Checking DCs and assertion-reason questions.

• General point, dot zero, double back

  Foot: general point, PQ · b = 0. Image: twice the foot minus P.

  Use it when: Foot of perpendicular and image questions.

• Cross the directions, dot the gap

  Skew distance = |(b₁ × b₂) · (a₂ − a₁)| ÷ |b₁ × b₂|.

  Use it when: Shortest distance.

• Parallel: cross the gap instead

  For parallel lines, cross b with the gap and divide by |b|.

  Use it when: Distance between parallel lines.

Three Dimensional Geometry: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section E style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Three Dimensional Geometry

The slips examiners see most often, and what to write instead.

• ✗Mistake: Reading direction ratios from a non-standard form

  Taking 3, 7, 4 from $\frac{x-5}{3}=\frac{4-y}{7}=\frac{2z-6}{4}$.

  Wrong working

  Direction ratios 3, 7, 4

  Right working

  Direction ratios 3, −7, 2 (after rewriting as (y − 4)/(−7) and (z − 3)/2)

  ✓Do this: Convert to standard form first.

  Usually costs: Usually the whole question.

• ✗Mistake: Using the skew formula for parallel lines

  $\overrightarrow{b_1}\times \overrightarrow{b_2}=\vec{0}$ makes the formula 0/0.

  ✓Do this: Check for parallel directions first; then use |b × (a₂ − a₁)| / |b|.

  Usually costs: 2-3 marks.

• ✗Mistake: Forgetting the modulus in the angle formula

  Getting an obtuse angle for lines.

  ✓Do this: Take the absolute value of the dot product to get the acute angle between lines.

  Usually costs: ½ to 1 mark.

• ✗Mistake: Dotting with the wrong vector for the foot

  Using PQ · (position vector) instead of PQ · (direction of the line).

  ✓Do this: PQ must be perpendicular to the direction ratios of the line.

  Usually costs: 3-4 marks.

• ✗Mistake: Not checking the third equation for intersection

  Solving two equations and assuming the lines meet.

  ✓Do this: Substitute λ and μ in the third equation; if it fails, the lines are skew.

  Usually costs: 1-2 marks.

NCERT exercise guide: Three Dimensional Geometry

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Three Dimensional Geometry: previous year questions

• Asked in 8 of the 8 board papers we hold (2023 to 2026), on average 8.3 marks per paper; 31 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Three Dimensional Geometry

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

Every Three Dimensional Geometry question, by year

Practice questions on Three Dimensional Geometry

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQBoard level

   The direction ratios of the line $\frac{x-2}{3}=\frac{1-y}{2}=\frac{z+4}{5}$ are:
   (a) 3, 2, 5
   (b) 3, −2, 5
   (c) −3, 2, 5
   (d) 2, −1, 4

   Hint

   Rewrite (1 − y)/2 in standard form.

   Answer

   (b) 3, −2, 5

2. Q2 · 1 mark · MCQBoard level

   If the direction cosines of a line are k, k, k, then:
   (a) k > 0
   (b) 0 < k < 1
   (c) k = 1
   (d) $k=\pm \frac{1}{\sqrt{3}}$

   Hint

   $3k^2=1$.

   Answer

   (d)

3. Q3 · 1 mark · MCQBoard level

   The distance of the point (3, 4, 5) from the x-axis is:
   (a) 3
   (b) 5
   (c) √41
   (d) √34

   Hint

   Use the other two coordinates.

   Answer

   (c) √(16 + 25) = √41

4. Q4 · 1 mark · Assertion-reasonBoard level

   Assertion (A): Lines with direction ratios 1, −2, 3 and 2, 1, 0 are perpendicular.
   Reason (R): Lines with direction ratios a₁, b₁, c₁ and a₂, b₂, c₂ are perpendicular if a₁a₂ + b₁b₂ + c₁c₂ = 0.
   (a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.

   Hint

   2 − 2 + 0.

   Answer

   (a)

5. Q5 · 2 marks · Very short answerHarder board level

   Show that the line through A(2, 1, 0) and B(4, 3, 1) is perpendicular to the line through C(1, 0, 3) and D(3, −1, 1).

   Hint

   Direction ratios from the points.

   Answer

   AB: 2, 2, 1; CD: 2, −1, −2. 4 − 2 − 2 = 0, so they are perpendicular.

6. Q6 · 2 marks · Very short answerHarder board level

   The Cartesian equation of a line is $\frac{x-3}{2}=\frac{y+1}{-3}=\frac{z-4}{6}$. Write its vector form and its direction cosines.

   Hint

   Magnitude of (2, −3, 6) is 7.

   Answer

   $\vec{r}=(3\hat{i}-\hat{j}+4\hat{k})+\lambda (2\hat{i}-3\hat{j}+6\hat{k})$; direction cosines $\frac{2}{7},-\frac{3}{7},\frac{6}{7}$.

7. Q7 · 3 marks · Short answerHarder board level

   Find the coordinates of the point where the line through A(2, −1, 3) and B(4, 3, −1) crosses the XY-plane (z = 0).

   Hint

   General point (2 + 2t, −1 + 4t, 3 − 4t).

   Answer

   z = 0 gives $t=\frac{3}{4}$; point $\left(\frac{7}{2},2,0\right)$.

8. Q8 · 3 marks · Short answerHarder board level

   Find the shortest distance between $\vec{r}=(2\hat{i}-\hat{k})+\lambda (\hat{i}+\hat{j}+\hat{k})$ and $\vec{r}=(\hat{i}+\hat{j})+\mu (\hat{i}-\hat{j}+2\hat{k})$.

   Hint

   b₁ × b₂ = 3i − j − 2k.

   Answer

   $|\overrightarrow{b_1}\times \overrightarrow{b_2}|=\sqrt{14}$; $\overrightarrow{a_2}-\overrightarrow{a_1}=-\hat{i}+\hat{j}+\hat{k}$; dot $=-3-1-2=-6$; $d=\frac{6}{\sqrt{14}}=\frac{3\sqrt{14}}{7}$ units.

9. Q9 · 3 marks · Short answerHarder board level

   Find the foot of the perpendicular from the origin to the line $\frac{x-2}{1}=\frac{y-3}{-1}=\frac{z+4}{2}$ and the length of the perpendicular.

   Hint

   OQ · (1, −1, 2) = 0.

   Answer

   $Q(2+t,3-t,-4+2t)$; $2+t-3+t-8+4t=6t-9=0$, $t=\frac{3}{2}$. Foot $\left(\frac{7}{2},\frac{3}{2},-1\right)$; length $\sqrt{\frac{49}{4}+\frac{9}{4}+1}=\frac{\sqrt{62}}{2}$.

10. Q10 · 5 marks · Long answerHarder board level

    Find the points on the line $\frac{x-1}{2}=\frac{y+1}{-2}=\frac{z-2}{1}$ at a distance of $\sqrt{62}$ units from P(4, 0, 7).

    Hint

    Set PQ² = 62 for a general point Q.

    Answer

    $Q(1+2t,-1-2t,2+t)$; $P{Q}^2=(2t-3{)}^2+(2t+1{)}^2+(t-5{)}^2=9t^2-18t+35=62\Rightarrow t^2-2t-3=0\Rightarrow t=3$ or −1. Points (7, −7, 5) and (−1, 1, 1).

11. Q11 · 2 marks · Very short answerHOTS / competency

    Can a line make angles 30°, 45° and 60° with the x-, y- and z-axes? Justify.

    Hint

    Test l² + m² + n² = 1.

    Answer

    ${\cos }^{2}30°+{\cos }^{2}45°+{\cos }^{2}60°=\frac{3}{4}+\frac{1}{2}+\frac{1}{4}=\frac{3}{2}\ne 1$, so no such line exists.

12. Q12 · 3 marks · Short answerHOTS / competency

    Find p so that the lines $\vec{r}=(\hat{i}-\hat{k})+\lambda (\hat{i}+\hat{j}+2\hat{k})$ and $\vec{r}=(4\hat{i}-\hat{j}+p\hat{k})+\mu (2\hat{i}-2\hat{j}+3\hat{k})$ intersect.

    Hint

    Intersecting lines have shortest distance 0.

    Answer

    $\overrightarrow{b_1}\times \overrightarrow{b_2}=7\hat{i}+\hat{j}-4\hat{k}$; $\overrightarrow{a_2}-\overrightarrow{a_1}=3\hat{i}-\hat{j}+(p+1)\hat{k}$; dot $=21-1-4p-4=16-4p=0$, so p = 4.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.What is l² + m² + n² for any line?Show answer

   1

2. 2.Direction cosines of the y-axis?Show answer

   0, 1, 0

3. 3.Vector equation of a line through a parallel to b?Show answer

   $\vec{r}=\vec{a}+\lambda \vec{b}$

4. 4.Condition for two lines to be perpendicular?Show answer

   a₁a₂ + b₁b₂ + c₁c₂ = 0

5. 5.What are skew lines?Show answer

   Lines that are neither parallel nor intersecting.

6. 6.Shortest distance if the lines intersect?Show answer

   0

7. 7.Image of P if the foot of the perpendicular is Q?Show answer

   2Q − P

8. 8.Distance of (a, b, c) from the y-axis?Show answer

   √(a² + c²)