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# NCERT Solutions for Class 12 maths chapter 11

# NCERT Solutions for Class 12 maths chapter 11

NCERT Solutions for Class 12 Maths Chapter 11 Exercise 11.1, 11.2, 11.3 & Miscellaneous exercises of Three dimensional geometry (3D) in PDF form to free download. * NCERT solutions*, Sample question papers, Assignments, test papers based on different difficulty levels, latest

*for the current academic year 2017-18.*

**CBSE syllabus**## NCERT Solutions for Class 12 Maths Chapter 11

**Click Here to NCERT Solutions Class 12 Maths**

### Class 12 Maths Solutions – Three Dimensional Geometry (3D)

#### NCERT Chapter to study online and answers given in the end of ncert books.

#### These books are very good for revision and more practice. These book are also confined to NCERT Syllabus.

#### Assignments for practice

**Level 1 Test 1 **

**Level 2 Test 1**

#### Previous Years Important Questions

- Find the Cartesian and Vector equations of the line which passes through the point (-2, 4, -5) and parallel to the line given by (x+3)/3 = (y-4)/5 = (8-z)/-6. [CBSE Sample Paper 2017]
- If the vectors
**p**= a**i**+**j**+**k**,**q**=**i**+ b**j**+**k**and**r**=**i**+**j**+ c**k**are co-planar, then for a, b, c ≠ 1, show that 1/(1-a) + 1/(1-b) + 1/(1-c) = 1. [CBSE Sample Paper 2017] - A plane meets the coordinate axes in A, B and C such that the centroid of triangle ABC is the point (α, β, γ). Show that the equation of the plane is x/α + y/β + z/γ = 3. [CBSE Sample Paper 2017]
- Define skew lines. Using only vector approach, find the shortest distance between the following two skew lines:
**r**= (8 + 3m)**i**– (9 + 16m)**j**+ (10 + 7m)**k**and**r**= 15**i**+ 29**j**+ 5**k**+ n(3**i**+ 8**j**– 5**k**). [CBSE Sample Paper 2017] - Find the vector equation of the line passing through the point A(1, 2, –1) and parallel to the line 5x – 25 = 14 – 7y = 35z. [Delhi 2017]
- Find the vector and Cartesian equations of a line passing through (1, 2, –4) and perpendicular to the two lines (x – 8)/3 = (y + 19)/-16 = (z – 10)/7 and (x – 15)/3 = (y – 29)/8 = (z – 5)/-5. [Delhi 2017]
- Find the vector equation of a plane which is at a distance of 5 units from the origin and its normal vector is 2
**i**– 3**j**+ 6**k**. [Delhi 2016] - Show that the vectors
**a**,**b**, and**c**are co-planar if**a + b + c**and**c + a**are co-planar. [Delhi 2016] - Find the coordinate of the point P where the line through A(3, – 4, –5) and B (2, –3, 1) crosses the plane passing through three points L(2, 2, 1), M(3, 0, 1) and N(4, –1, 0). Also, find the ratio in which P divides the line segment AB. [Delhi 2016]
- Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x – y + z = 0. Then find the distance of plane thus obtained from the point A(1, 3, 6). [Delhi 2015C]