CBSE Class 12 Mathematics Three Dimensional Geometry VBQs Set C

Read and download the CBSE Class 12 Mathematics Three Dimensional Geometry VBQs Set C. Designed for the 2025-26 academic year, these Value Based Questions (VBQs) are important for Class 12 Mathematics students to understand moral reasoning and life skills. Our expert teachers have created these chapter-wise resources to align with the latest CBSE , NCERT, and KVS examination patterns.

VBQ for Class 12 Mathematics Chapter 11 Three Dimensional Geometry

For Class 12 students, Value Based Questions for Chapter 11 Three Dimensional Geometry help to apply textbook concepts to real-world application. These competency-based questions with detailed answers help in scoring high marks in Class 12 while building a strong ethical foundation.

Chapter 11 Three Dimensional Geometry Class 12 Mathematics VBQ Questions with Answers

Short Answer Questions–I

Question. Find the coordinates of the point where the line \( \frac{x+1}{2} = \frac{y+2}{3} = \frac{z+3}{4} \) meets the plane \( x + y + 4z = 6 \).
Answer: (1, 1, 1)

Question. If the line drawn from the point (–2, –1, –3) meets a plane at right angle at the point (1, –3, 3), find the equation of the plane.
Answer: \( 3x - 2y + 6z - 27 = 0 \)

Question. Find the distance of the point whose position vector is \( (2\hat{i} + \hat{j} - \hat{k}) \) from the plane \( \vec{r} \cdot (\hat{i} - 2\hat{j} + 4\hat{k}) = 9 \).
Answer: \( \frac{13}{\sqrt{21}} \)

Question. Write the unit vector normal to the plane \( x + 2y + 3z - 6 = 0 \).
Answer: \( \frac{1}{\sqrt{14}}\hat{i} + \frac{2}{\sqrt{14}}\hat{j} + \frac{3}{\sqrt{14}}\hat{k} \)

Question. Find the vector equation of a plane which is at a distance of 5 units from the origin and its normal vector is \( 2\hat{i} - 3\hat{j} + 6\hat{k} \).
Answer: \( \vec{r} \cdot \left( \frac{2}{7}\hat{i} - \frac{3}{7}\hat{j} + \frac{6}{7}\hat{k} \right) = 5 \)

Question. Find the equation of the plane passing through the line of intersection of the planes \( 2x + 2y - 3z = 7 \) and \( 2x + 5y + 3z = 9 \) the point (2, 1, 3).
Answer: \( 38x + 68y + 3z = 153 \)

Question. Find the angle between the planes, whose vector equations are \( \vec{r} \cdot (2\hat{i} + 2\hat{j} - 3\hat{k}) = 5 \) and \( \vec{r} \cdot (3\hat{i} - 3\hat{j} + 5\hat{k}) = 3 \).
Answer: \( \cos^{-1}\left( \frac{15}{\sqrt{731}} \right) \)

Short Answer Questions–II

Question. Find the shortest distance between the following pair of skew lines: \( \frac{x-1}{2} = \frac{2-y}{3} = \frac{z+1}{4} , \frac{x+2}{1} = \frac{y-3}{2} = \frac{z}{3} \).
Answer: \( \frac{42}{\sqrt{390}} \) units

Question. Show that the lines \( \frac{x-1}{3} = \frac{y-1}{-1} = \frac{z+1}{0} \) and \( \frac{x-4}{2} = \frac{y}{0} = \frac{z+1}{3} \) intersect. Find their point of intersection.
Answer: (4, 0, –1)

Question. Find the coordinates of the foot of perpendicular drawn from the point A(–1, 8, 4) to the line joining the points \( B(0, -1, 3) \) and \( C(2, -3, -1) \). Hence find the image of the point A in the line BC.
Answer: (–2, 1, 7); (–3, –6, 10)

Question. Find the equation of plane passing through the points \( A(3, 2, 1), B(4, 2, -2) \) and \( C(6, 5, -1) \) and hence find the value of \(\lambda\) for which \( A(3, 2, 1), B(4, 2, -2), C(6, 5, -1) \) and \( D(\lambda, 5, 5) \) are coplanar.
Answer: \( 9x - 7y + 3z - 16 = 0 \); \( \lambda = 4 \)

Question. Prove that the line through \( A(0, -1, -1) \) and \( B(4, 5, 1) \) intersects the line through \( C(3, 9, 4) \) and \( D(-4, 4, 4) \).
Answer: The lines intersect at (2, 2, 0).

Question. Show that the following two lines are coplanar: \( \frac{x-a+d}{\alpha-\delta} = \frac{y-a}{\alpha} = \frac{z-a-d}{\alpha+\delta} \) and \( \frac{x-b+c}{\beta-\gamma} = \frac{y-b}{\beta} = \frac{z-b-c}{\beta+\gamma} \).
Answer: The lines are coplanar since the determinant condition for coplanarity is satisfied.

Question. Find the acute angle between the plane \( 5x - 4y + 7z - 13 = 0 \) and the y-axis.
Answer: \( \sin^{-1} \left( \frac{4}{3\sqrt{10}} \right) \)

Question. Find the equation of a plane which is at a distance of \( 3\sqrt{3} \) units from origin and the normal to which is equally inclined to the coordinate axes.
Answer: \( x + y + z = 9 \)

Question. Let \( P(3, 2, 6) \) be a point in the space and Q be a point on the line \( \vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \mu(-3\hat{i} + \hat{j} + 5\hat{k}) \), then find the value of \(\mu\) for which the vector \( \vec{PQ} \) is parallel to the plane \( x - 4y + 3z = 1 \).
Answer: \( \mu = \frac{1}{4} \)

Question. Find the vector and cartesian equations of the plane which bisects the line joining the points (3, –2, 1) and (1, 4, –3) at right angles.
Answer: \( \vec{r} \cdot (\hat{i} - 3\hat{j} + 2\hat{k}) + 3 = 0 \); \( x - 3y + 2z + 3 = 0 \)

Question. Find the distance of the point \( P(3, 4, 4) \) from the point, where the line joining the points \( A(3, -4, -5) \) and \( B(2, -3, 1) \) intersect the plane \( 2x + y + z = 7 \).
Answer: 7 units

Question. Find the distance of the point (1, –2, 3) from the plane \( x - y + z = 5 \) measured parallel to the line whose direction cosines are proportional to 2, 3, –6.
Answer: 1 unit

Question. Find the equation of the plane containing two parallel lines \( \frac{x-1}{2} = \frac{y+1}{-1} = \frac{z}{3} \) and \( \frac{x}{4} = \frac{y-2}{-2} = \frac{z+1}{6} \). Also, find if the plane thus obtained contains the line \( \frac{x-2}{3} = \frac{y-1}{1} = \frac{z-2}{5} \) or not.
Answer: \( 8x + y - 5z = 7 \); yes the plane contains the given line.

Question. Find the vector and cartesian equations of a line through the point (1, –1, 1) and perpendicular to the lines joining the points (4, 3, 2), (1, –1, 0) and (1, 2, –1), (2, 1, 1).
Answer: \( \frac{x-1}{10} = \frac{y+1}{-4} = \frac{z-1}{-7} \); \( \vec{r} = (\hat{i} - \hat{j} + \hat{k}) + \lambda(10\hat{i} - 4\hat{j} - 7\hat{k}) \)

Question. Find the shortest distance between the lines whose vector equations are \( \vec{r} = (1-t)\hat{i} + (t-2)\hat{j} + (3-2t)\hat{k} \) and \( \vec{r} = (s+1)\hat{i} + (2s-1)\hat{j} + (2s+1)\hat{k} \).
Answer: \( \frac{8}{\sqrt{29}} \) units

Question. Find the vector and cartesian equations of the line passing through the point P (1, 2, 3) and parallel to the planes \( \vec{r} \cdot (\hat{i} - \hat{j} + 2\hat{k}) = 5 \) and \( \vec{r} \cdot (3\hat{i} + \hat{j} + \hat{k}) = 6 \).
Answer: \( \vec{r} = (\hat{i} + \hat{j} + \hat{k}) + \lambda(-3\hat{i} + 5\hat{j} + 4\hat{k}) \); \( \frac{x-1}{-3} = \frac{y-2}{5} = \frac{z-3}{4} \)

Long Answer Questions

Question. Find the equation of the plane passing through the point P(1, 1, 1) and containing the line \( \vec{r} = (-3\hat{i} + \hat{j} + 5\hat{k}) + \lambda(3\hat{i} - \hat{j} - 5\hat{k}) \). Also, show that the plane contains the line \( \vec{r} = (-\hat{i} + 2\hat{j} + 5\hat{k}) + \mu(\hat{i} - 2\hat{j} - 5\hat{k}) \).
Answer: \( \vec{r} \cdot (\hat{i} - \hat{j} + \hat{k}) = 0 \)

Question. Find the vector and cartesian equations of the plane passing through the intersection of the planes \( \vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 6 \) and \( \vec{r} \cdot (2\hat{i} + 3\hat{j} + 4\hat{k}) = -5 \) and the point (1, 1, 1).
Answer: \( 20x + 23y + 26z - 69 = 0 \); \( \vec{r} \cdot (20\hat{i} + 23\hat{j} + 26\hat{k}) = 69 \)

Question. Find the value of k for which the following lines are perpendicular to each other: \( \frac{x+3}{k-5} = \frac{y-1}{1} = \frac{5-z}{-2k-1} \) and \( \frac{x+2}{-1} = \frac{2-y}{-k} = \frac{z}{5} \). Hence, find the equation of the plane containing the above lines.
Answer: \( k = -1 \); \( 4x + 31y + 7z = 54 \)

Question. Show that the lines: \( \vec{r} = \hat{i} + \hat{j} + \hat{k} + \lambda(\hat{i} - \hat{j} + \hat{k}) \) and \( \vec{r} = 4\hat{j} + 2\hat{k} + \mu(2\hat{i} - \hat{j} + 3\hat{k}) \) are coplanar. Also, find the equation of the plane containing these lines.
Answer: \( \vec{r} \cdot (-2\hat{i} - \hat{j} + \hat{k}) + 2 = 0 \)

Question. Show that the lines \( \vec{r} = (-3\hat{i} + \hat{j} + 5\hat{k}) + \lambda(3\hat{i} - \hat{j} - 5\hat{k}) \) and \( \vec{r} = (-\hat{i} + 2\hat{j} + 5\hat{k}) + \mu(\hat{i} - 2\hat{j} - 5\hat{k}) \) are co-planar. Also, find the equation of the plane containing these lines.
Answer: \( x - 2y + z = 0 \)

Question. Find the position vector of the foot of perpendicular and the perpendicular distance from the point P with position vector \( 2\hat{i} + 3\hat{j} + 4\hat{k} \) to the plane \( \vec{r} \cdot (2\hat{i} + \hat{j} + 3\hat{k}) - 26 = 0 \). Also, find image of P in the plane.
Answer: \( 3\hat{i} + \frac{7}{2}\hat{j} + \frac{11}{2}\hat{k} \); \( \sqrt{\frac{7}{2}} \) units; (4, 4, 7)

Question. Find the coordinates of the foot of perpendicular and perpendicular distance from the point P(4, 3, 2) to the plane \( x + 2y + 3z = 2 \). Also find the image of P in the plane.
Answer: (3, 1, –1); \( \sqrt{14} \) sq. units; (2, –1, –4)

Question. Find the equation of the plane which contains the line of intersection of the planes \( x + 2y + 3z - 4 = 0 \) and \( 2x + y - z + 5 = 0 \) and whose x-intercept is twice its z-intercept. Hence write the vector equation of a plane passing through the point (2, 3, –1) and parallel to the plane obtained above.
Answer: \( 13x + 14y + 11z = 0 \) or \( 7x + 11y + 14z - 15 = 0 \); \( \{ \vec{r} - (2\hat{i} + 3\hat{j} - \hat{k}) \} \cdot (7\hat{i} + 11\hat{j} + 14\hat{k}) = 0 \)

Question. Find the coordinate of the point where the line through (3, –4, –5) and (2, –3, 1) crosses the plane, passing through the points (2, 2, 1), (3, 0, 1) and (4, –1, 0).
Answer: (1, –2, 7)

Question. Show that lines \( \vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(3\hat{i} - \hat{j}) \) and \( \vec{r} = (4\hat{i} - \hat{k}) + \mu(2\hat{i} + 3\hat{k}) \) intersect. Also find their point of intersection.
Answer: (4, 0, –1)

Question. Find the distance between the point (7, 2, 4) and the plane determine by the points \( A(2, 5, -3), B(-2, -3, 5) \) and \( C(5, 3, -3) \).
Answer: 29 units

Question. Show that the lines \( \frac{x-2}{1} = \frac{y-2}{3} = \frac{z-3}{1} \) and \( \frac{x-2}{1} = \frac{y-3}{4} = \frac{z-4}{2} \) intersect. Also, find the co-ordinate of the point of intersection and equation of the plane containing the two lines.
Answer: (1, –1, 2); \( 2x - y + z - 5 = 0 \)

VBQs for Chapter 11 Three Dimensional Geometry Class 12 Mathematics

Students can now access the Value-Based Questions (VBQs) for Chapter 11 Three Dimensional Geometry as per the latest CBSE syllabus. These questions have been designed to help Class 12 students understand the moral and practical lessons of the chapter. You should practicing these solved answers to improve improve your analytical skills and get more marks in your Mathematics school exams.

Expert-Approved Chapter 11 Three Dimensional Geometry Value-Based Questions & Answers

Our teachers have followed the NCERT book for Class 12 Mathematics to create these important solved questions. After solving the exercises given above, you should also refer to our NCERT solutions for Class 12 Mathematics and read the answers prepared by our teachers.

Improve your Mathematics Scores

Daily practice of these Class 12 Mathematics value-based problems will make your concepts better and to help you further we have provided more study materials for Chapter 11 Three Dimensional Geometry on studiestoday.com. By learning these ethical and value driven topics you will easily get better marks and also also understand the real-life application of Mathematics.

Where can I find 2025-26 CBSE Value Based Questions (VBQs) for Class 12 Mathematics Chapter Chapter 11 Three Dimensional Geometry?

The latest collection of Value Based Questions for Class 12 Mathematics Chapter Chapter 11 Three Dimensional Geometry is available for free on StudiesToday.com. These questions are as per 2026 academic session to help students develop analytical and ethical reasoning skills.

Are answers provided for Class 12 Mathematics Chapter Chapter 11 Three Dimensional Geometry VBQs?

Yes, all our Mathematics VBQs for Chapter Chapter 11 Three Dimensional Geometry come with detailed model answers which help students to integrate factual knowledge with value-based insights to get high marks.

What is the importance of solving VBQs for Class 12 Chapter Chapter 11 Three Dimensional Geometry Mathematics?

VBQs are important as they test student's ability to relate Mathematics concepts to real-life situations. For Chapter Chapter 11 Three Dimensional Geometry these questions are as per the latest competency-based education goals.

How many marks are usually allocated to VBQs in the CBSE Mathematics paper?

In the current CBSE pattern for Class 12 Mathematics, Chapter 11 Three Dimensional Geometry Value Based or Case-Based questions typically carry 3 to 5 marks.

Can I download Mathematics Chapter Chapter 11 Three Dimensional Geometry VBQs in PDF for free?

Yes, you can download Class 12 Mathematics Chapter Chapter 11 Three Dimensional Geometry VBQs in a mobile-friendly PDF format for free.