Three Dimensional Geometry Class 12 (3D Geometry Class 12)

Master Three Dimensional Geometry Class 12 (3D Geometry Class 12) with comprehensive NCERT Solutions, Practice Questions, MCQs, Sample Papers, Case Based Questions, and Video lessons.

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NCERT Solutions

Three Dimensional Geometry Class 12 (3D Geometry Class 12) – NCERT Solutions

Each question below opens its complete step-by-step Teachoo solution.

Ex 11.1

5 questions

Ex 11.1, 1

If a line makes angles $90^{\circ}, 135^{\circ}, 45^{\circ}$ with the $x, y$ and $z$-axes respectively, find its direction cosines.

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Ex 11.1, 2

Find the direction cosines of a line which makes equal angles with the coordinate axes.

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Ex 11.1, 3

If a line has the direction ratios -18, 12, -4, then what are its direction cosines?

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Ex 11.1, 4

Show that the points $(2,3,4),(-1,-2,1),(5,8,7)$ are collinear.

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Ex 11.1, 5

Find the direction cosines of the sides of the triangle whose vertices are $(3,5,-4),(-1,1,2)$ and $(-5,-5,-2)$.

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Ex 11.2

17 questions

Ex 11.2, 1

Show that the three lines with direction cosines
$$
\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13} ; \frac{4}{13}, \frac{12}{13}, \frac{3}{13} ; \frac{3}{13}, \frac{-4}{13}, \frac{12}{13} \text { are mutually perpendicular. }
$$

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Ex 11.2, 2

Show that the line through the points $(1,-1,2),(3,4,-2)$ is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).

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Ex 11.2, 3

Show that the line through the points $(4,7,8),(2,3,4)$ is parallel to the line through the points (-1, -2, 1), (1, 2, 5).

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Ex 11.2, 4

Find the equation of the line which passes through the point $(1,2,3)$ and is parallel to the vector $3 \hat{i}+2 \hat{j}-2 \hat{k}$.

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Ex 11.2, 5

Find the equation of the line in vector and in cartesian form that passes through the point with position vector $2 \hat{i}-j+4 \hat{k}$ and is in the direction $\hat{i}+2 \hat{j}-\hat{k}$.

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Ex 11.2, 6

Find the cartesian equation of the line which passes through the point $(-2,4,-5)$ and parallel to the line given by $\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$.

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Ex 11.2, 7

The cartesian equation of a line is $\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}$. Write its vector form.

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Ex 11.2, 8 (i)

Find the angle between the following pairs of lines:
(i)
$$
\begin{aligned}
\vec{r} & =2 \hat{i}-5 \hat{j}+\hat{k}+\lambda(3 \hat{i}+2 \hat{j}+6 \hat{k}) \text { and } \\
\vec{r} & =7 \hat{i}-6 \hat{k}+\mu(\hat{i}+2 \hat{j}+2 \hat{k})
\end{aligned}
$$

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Ex 11.2, 8 (ii)

Find the angle between the following pairs of lines:
(ii)
&\begin{aligned}
& \vec{r}=3 \hat{i}+\hat{j}-2 \hat{k}+\lambda(\hat{i}-\hat{j}-2 \hat{k}) \text { and } \\
& \vec{r}=2 \hat{i}-\hat{j}-56 \hat{k}+\mu(3 \hat{i}-5 \hat{j}-4 \hat{k})
\end{aligned}

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Ex 11.2, 9 (i)

Find the angle between the following pair of lines:
(i) $\frac{x-2}{2}=\frac{y-1}{5}=\frac{z+3}{-3}$ and $\frac{x+2}{-1}=\frac{y-4}{8}=\frac{z-5}{4}$

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Ex 11.2, 9 (ii)

Find the angle between the following pair of lines:
(ii) $\frac{x}{2}=\frac{y}{2}=\frac{z}{1}$ and $\frac{x-5}{4}=\frac{y-2}{1}=\frac{z-3}{8}$

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Ex 11.2, 10

Find the values of $p$ so that the lines $\frac{1-x}{3}=\frac{7 y-14}{2 p}=\frac{z-3}{2}$ and $\frac{7-7 x}{3 p}=\frac{y-5}{1}=\frac{6-z}{5}$ are at right angles.

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Ex 11.2, 11

Show that the lines $\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}$ and $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ are perpendicular to each other.

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Ex 11.2, 12

Find the shortest distance between the lines
$$
\begin{aligned}
& \vec{r}=(\hat{i}+2 \hat{j}+\hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k}) \text { and } \\
& \vec{r}=2 \hat{i}-\hat{j}-\hat{k}+\mu(2 \hat{i}+\hat{j}+2 \hat{k})
\end{aligned}
$$

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Ex 11.2, 13

Find the shortest distance between the lines
$$
\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1} \text { and } \frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}
$$

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Ex 11.2, 14

Find the shortest distance between the lines whose vector equations are
$$
\begin{aligned}
& \vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}-3 \hat{j}+2 \hat{k}) \\
& \text { and } \vec{r}=4 \hat{i}+5 \hat{j}+6 \hat{k}+\mu(2 \hat{i}+3 \hat{j}+\hat{k})
\end{aligned}
$$

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Ex 11.2, 15

Find the shortest distance between the lines whose vector equations are
$$
\begin{aligned}
\vec{r} & =(1-t) \hat{i}+(t-2) \hat{j}+(3-2 t) \hat{k} \text { and } \\
\vec{r} & =(s+1) \hat{i}+(2 s-1) \hat{j}-(2 s+1) \hat{k}
\end{aligned}
$$

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Examples

10 questions

Example 1

If a line makes angle $90^{\circ}, 60^{\circ}$ and $30^{\circ}$ with the positive direction of $x, y$ and $z$-axis respectively, find its direction cosines.

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Example, 2

If a line has direction ratios 2, -1, -2, determine its direction cosines

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Example, 3

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

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Example, 5

Show that the points $\mathrm{A}(2,3,-4), \mathrm{B}(1,-2,3)$ and $\mathrm{C}(3,8,-11)$ are collinear.

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Example, 6

Find the vector and the Cartesian equations of the line through the point $(5,2,-4)$ and which is parallel to the vector $3 \hat{i}+2 \hat{j}-8 \hat{k}$.

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Example, 7

Find the angle between the pair of lines given by
$$
\vec{r}=3 \hat{i}+2 \hat{j}-4 \hat{k}+\lambda(\hat{i}+2 \hat{j}+2 \hat{k})
$$
and
$$
\vec{r}=5 \hat{i}-2 \hat{j}+\mu(3 \hat{i}+2 \hat{j}+6 \hat{k})
$$

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Example 8

Find the angle between the pair of lines
and
$$
\begin{aligned}
& \frac{x+3}{3}=\frac{y-1}{5}=\frac{z+3}{4} \\
& \frac{x+1}{1}=\frac{y-4}{1}=\frac{z-5}{2}
\end{aligned}
$$

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Example 9

Find the shortest distance between the lines $l_1$ and $l_2$ whose vector equations are
$$
\vec{r}=\hat{i}+\hat{j}+\lambda(2 \hat{i}-\hat{j}+\hat{k}) \tag{1}
$$
and
$$
\vec{r}=2 \hat{i}+\hat{j}-\hat{k}+\mu(3 \hat{i}-5 \hat{j}+2 \hat{k}) \tag{2}
$$

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Example 10

Find the distance between the lines $l_1$ and $l_2$ given by
and
$$
\begin{aligned}
& \vec{r}=\hat{i}+2 \hat{j}-4 \hat{k}+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k}) \\
& \vec{r}=3 \hat{i}+3 \hat{j}-5 \hat{k}+\mu(2 \hat{i}+3 \hat{j}+6 \hat{k})
\end{aligned}
$$

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Miscellaneous

5 questions

Misc 1

Find the angle between the lines whose direction ratios are $a, b, c$ and $b-c, c-a, a-b$.

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Misc 2

Find the equation of a line parallel to $x$-axis and passing through the origin.

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Misc 3

If the lines $\frac{x-1}{-3}=\frac{y-2}{2 k}=\frac{z-3}{2}$ and $\frac{x-1}{3 k}=\frac{y-1}{1}=\frac{z-6}{-5}$ are perpendicular, find the value of $k$.

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Misc 4

Find the shortest distance between lines $\vec{r}=6 \hat{i}+2 \hat{j}+2 \hat{k}+\lambda(\hat{i}-2 \hat{j}+2 \hat{k})$ and $\vec{r}=-4 \hat{i}-\hat{k}+\mu(3 \hat{i}-2 \hat{j}-2 \hat{k})$.

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Misc 5

Find the vector equation of the line passing through the point $(1,2,-4)$ and perpendicular to the two lines:
$$
\frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7} \text { and } \frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5} .
$$

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Why Learn This With Teachoo?

Three-Dimensional Geometry uses vectors and coordinates to study lines and planes in space. Students learn direction cosines and ratios, vector and Cartesian equations, angles, intersections, coplanarity and distances. Teachoo provides NCERT solutions, examples, miscellaneous questions and concept-wise methods for lines, planes and spatial relationships.

Direction and equations of a line

Direction cosines l, m and n are cosines of the angles a line makes with the positive coordinate axes and satisfy l²+m²+n²=1. Direction ratios are proportional triples; dividing them by their magnitude gives direction cosines, with signs determined by orientation.

A line through position vector a and parallel to vector b has equation r=a+λb. In Cartesian form, (x−x₁)/a=(y−y₁)/b=(z−z₁)/c, where a,b,c are direction ratios. Lines may intersect, be parallel or be skew. The angle between lines is calculated from their direction vectors.

Equations of a plane

A plane through a point with normal vector n has vector equation (r−a)·n=0 and Cartesian form A(x−x₁)+B(y−y₁)+C(z−z₁)=0. The general form Ax+By+Cz+D=0 has normal vector (A,B,C).

Planes can be described using a normal and point, three non-collinear points, intercepts or relationships to other lines and planes. Angles between planes come from their normals. A line-plane angle is complementary to the angle between the line direction and plane normal.

Distances and relationships

The distance from (x₁,y₁,z₁) to Ax+By+Cz+D=0 is |Ax₁+By₁+Cz₁+D|/√(A²+B²+C²). Parallel-plane distance follows after coefficients are made identical. Students also calculate distance between parallel or skew lines and test coplanarity using scalar triple products or intersection conditions.

Topics and resources on Teachoo

  • NCERT exercises, examples and miscellaneous solutions;

  • direction cosines and direction ratios;

  • vector and Cartesian line equations;

  • angles, intersection and coplanarity of lines;

  • plane equations in different forms;

  • angles between lines and planes;

  • point-to-plane and parallel-plane distance;

  • shortest distances involving lines;

  • board and entrance-oriented spatial questions.

Learning outcomes

Students should be able to convert between vector and Cartesian forms, create equations from geometric conditions and calculate angles and distances. They should classify spatial relationships and use vector products to test perpendicularity, parallelism or coplanarity.

Board and entrance-exam preparation

Extract direction and normal vectors before calculating. Sketch the relationship even if the 3D drawing is rough. Use absolute values in distances and state whether the requested angle is between lines, normals or a line and plane. Substitute an intersection point back into every equation.

Common mistakes to avoid

Do not confuse direction ratios with direction cosines. A plane’s coefficients give a normal, not a direction lying in the plane. Skew lines are non-parallel and non-intersecting. Distance formulas require compatible standard forms. The angle between a line and plane is not directly the angle between line direction and plane normal.

Deeper reasoning and concept connections

In Three-Dimensional Geometry, fluency means more than repeating a procedure. Students should be able to recognise the underlying structure when the numbers, diagram, wording or orientation changes. A useful routine is: identify the mathematical objects, list the known and unknown quantities, state the governing property, carry out the steps and verify that every condition has been used.

Look for connections within the chapter as well. A definition usually leads to a representation; the representation reveals a pattern; and the pattern supports a rule or calculation. Explaining this chain improves retention and helps with case-based questions. It also prevents the common mistake of selecting a formula simply because its symbols resemble the numbers in the question.

How to solve unfamiliar and competency-based questions

Use a five-step response: interpret, represent, select, solve and verify. Interpret the wording; represent the information; select a definition, property or formula; solve without skipping the logical step; and verify through substitution, estimation, measurement or an alternative representation. This routine works for direct exercises as well as case-based questions.

If information appears unnecessary, ask whether it establishes a hidden condition. If information is missing, state what cannot be determined instead of inventing a value. In written answers, name the rule being used. Clear reasoning helps a teacher award method marks and also makes the page easier for a student—or an AI answer system—to retrieve for the precise doubt being asked.

What complete mastery looks like

For Three-Dimensional Geometry, a student should be able to define the central ideas in simple language, recognise them in different representations, solve routine questions accurately and explain the method used. They should also be able to correct a flawed solution, create an example satisfying given conditions and combine two ideas from the chapter in one problem. A reliable mastery test is to solve one direct question, one application question and one reasoning question without looking at notes, then explain all three aloud or in writing.

Keep a compact error log with four labels: concept, interpretation, calculation and presentation. Reattempt each error after a gap instead of rereading the answer immediately. Improvement comes from correcting the decision that caused the mistake, not from repeating questions whose method is already known.

Additional frequently asked questions

What should a student know before starting Three-Dimensional Geometry?

Revise the definitions, number operations, diagrams or notation used at the beginning of the chapter. The prerequisite list should be short: if an earlier skill blocks progress, repair that skill with two or three focused questions and return to the chapter.

How can a student check an answer in Three-Dimensional Geometry?

Use an independent check whenever possible: substitute the result, reverse the operation, estimate its size, compare it with the figure, test a simpler case or solve using another representation. A check should examine the mathematical condition, not merely repeat the same arithmetic.

How many questions are enough for strong preparation?

There is no fixed number. Stop counting questions and track coverage: every concept, every standard method, at least one mixed problem, one competency-based problem and every previously incorrect type should be solved independently. Ten varied, analysed questions are more valuable than fifty copied solutions.

How should Teachoo solutions be used without becoming dependent on them?

Attempt the question first and mark the exact step where progress stops. Read only enough of the solution to repair that step, close it and restart the question. Finally, solve a similar problem without help. This turns a solution into feedback rather than a substitute for thinking.

Frequently asked questions

What is the normal vector of Ax+By+Cz+D=0?

It is (A,B,C).

What are skew lines?

They are lines in space that are neither parallel nor intersecting.

How are direction cosines obtained from direction ratios?

Divide each direction ratio by the square root of the sum of their squares, using the correct orientation signs.