NCERT Solutions for Class 12 Maths Miscellaneous Exercise Three Dimensional Geometry

NCERT Solutions for Class 12 Maths Miscellaneous Exercise: Miscellaneous exercises are often not taken in to consideration by the students. But, the fact is that miscellaneous exercises are important to solve for solving the High Order Thinking questions. The chapter 11, Three Dimensional Geometry is an important chapter that is designed by the subject experts according to the latest CBSE guidelines. The NCERT solutions of miscellaneous exercise is available at Embibe for!

Latest Update and NCERT 12 Maths Solutions for

• CBSE Class 12 Maths Exam Date – on 11th March 2023
• CBSE Class 12 Maths Exam Notes
• CBSE Class 12 Maths Sample Papers
• CBSE Class 12 Maths Mock Test Series
• List of Best Books for Maths CBSE Class 12

The NCERT solutions for Maths Miscellaneous Chapter 11 Class 12 is created by the best experts and hence students can rely on them with complete confidence. Students can sign up for at Embibe and attend all the mock tests, refer the study materials and understand the chapter. In this way, students will score better. Read along to find out more!

NCERT Solutions for Class 12 Maths Miscellaneous Exercise: Important Topics

The NCERT Solutions for Miscellaneous Exercise Chapter 11 Class 12 is important for the students to be thorough with. This is because the questions are asked from the NCERT books in the board exam and students who are thorough with the questions can score great in the exam. Given below are the important exercise topics of the chapter, Three Dimentional Geometry for an idea.

NCERT Class 12 Maths Chapter 11: Points To Remember

Students can refer to the chapter 11 class 12 important formulas to have a clear idea on the chapter from the points given below:

1. Distance formula:

(i) If P(x1, y1, z1)Px1, y1, z1 and Q(x2, y2, z2)Qx2, y2, z2 are two points in space, then PQ=(x2 − x1)2 + (y2 − y1)2 + (z2 −z1)2−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√PQ=x2 – x12 + y2 – y12 + z2 -z12

(ii) The distance of a point P (x, y, z)P (x, y, z) from the origin OO is given by OP=x2  +  y2  +  z2−−−−−−−−−−−−−−√OP=x2  +  y2  +  z2.

(iii) The distances of point P (x, y, z)P (x, y, z) from x, yx, y and zz axes are y2  +  z2−−−−−−−−√,  z2  +  x2−−−−−−−−√  and x2  +  y2−−−−−−−−√y2  +  z2,  z2  +  x2  and x2  +  y2 respectively.

2. Section formula:

P(x1, y1, z1)Px1, y1, z1 and Q(x2, y2, z2)Qx2, y2, z2 are two points

(i) Internal division:

The coordinates of a point dividing PQPQ internally in the ratio m:nm:n are (mx2 + nx1m + n,  my2 + ny1m + n,  mz2 + nz1m + n)mx2 + nx1m + n,  my2 + ny1m + n,  mz2 + nz1m + n

(ii) External division:

The coordinates of a point dividing PQPQ externally in the ratio m:n,m:n, then its coordinates are (mx2−nx1m−n,  my2−ny1m−n,  mz2−nz1m−n)mx2-nx1m-n,  my2-ny1m-n,  mz2-nz1m-n

(iii) The coordinates of the mid-point of PQPQ are (x1 + x22, y1 + y22, z1 + z22)x1 + x22, y1 + y22, z1 + z22

(iv) The line segment joining P(x1, y1, z1)Px1, y1, z1 and Q(x2, y2, z2)Qx2, y2, z2 is divided by

(a) YZYZ-plane in the ratio ‒x1 :x2‒x1 :x2

(b) XZXZ-plane in the ratio ‒y1:y2‒y1:y2

(c) XYXY-plane in the ratio ‒z1:z2‒z1:z2

(v) Centroid of a triangle:

The coordinates of the centroid of the triangle formed by the points (x1, y1, z1)x1, y1, z1, (x2, y2, z2)(x2, y2, z2) and (x3, y3, z3)x3, y3, z3 are (x1 + x2 + x33, y1 + y2 + y33, z1 + z2 + z33)x1 + x2 + x33, y1 + y2 + y33, z1 + z2 + z33

(vi) Centroid of a tetrahedron:

The coordinates of the centroid of the tetrahedron formed by the points (x1, y1, z1), (x2, y2, z2), (x3, y3, z3)x1, y1, z1, x2, y2, z2, x3, y3, z3 and (x4, y4, z4)x4, y4, z4 are (x1 + x2 + x3 + x44, y1 + y2 + y3 + y44, z1 + z2 + z3 +z44)x1 + x2 + x3 + x44, y1 + y2 + y3 + y44, z1 + z2 + z3 +z44

3. Direction cosines and direction ratios:

(i) Direction cosines:

(a) If a directed line segment OPOP makes angles α, β, γα, β, γ with OX, OYOX, OY and OZOZ respectively, then cos α, cos β, cos γcos α, cos β, cos γ are the direction cosines of OPOP and are generally denoted by l, m, n.l, m, n. Thus, we have l=cos α, m=cos β, n=cos γl=cos α, m=cos β, n=cos γ and direction cosines of POPO are ‒ l, ‒m, ‒n.‒ l, ‒m, ‒n.

(b) If OP = rOP = r and the coordinates of PP are (x, y, z),(x, y, z), then x=lr, y=mr,z=nr.x=lr, y=mr,z=nr.

(c) If l, m, nl, m, n are direction cosines of a vector r→r→, then r→ = ∣∣r→∣∣ (l iˆ + m jˆ + n kˆ) where, rˆ = l iˆ + m jˆ + n kˆr→ = |r→| (l i^ + m j^ + n k^) where, r^ = l i^ + m j^ + n k^ and, l2+m2+n2=1l2+m2+n2=1.

(d) ∣∣∣r→∣∣∣=Sum of the squares of projections of r→ on the coordinate axes−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√|r→|=Sum of the squares of projections of r→ on the coordinate axes

(ii) Direction ratios:

If l, m, nl, m, n are direction cosines of a vector r→r→ and a, b, ca, b, c are three numbers such that la=mb=ncla=mb=nc, then we say that the direction ratios of r→r→ are proportional to a, b, ca, b, c.

(iii) Angle between two lines for the given direction cosines:

If θθ is the angle between two lines having direction cosines l1, m1, n1l1, m1, n1 and l2, m2, n2l2, m2, n2 then,

(a) cos θ= l1l2+m1m2+n1n2cos θ= l1l2+m1m2+n1n2

(b) Lines are parallel, if l1l2 = m1m2 = n1n2l1l2 = m1m2 = n1n2

(c) Lines are perpendicular, if l1l2+m1m2+n1n2=0l1l2+m1m2+n1n2=0

(iv) Angle between two lines for the given direction ratios:

(a) If θθ is the angle between two lines whose direction ratios are proportional to a1, b1, c1a1, b1, c1 and a2, b2, c2a2, b2, c2 respectively, then the angle θθ between them is given by cos θ=a1a2 + b1b2 + c1c2a21 + b21 + c21√  a22 + b22 + c22√cos θ=a1a2 + b1b2 + c1c2a12 + b12 + c12  a22 + b22 + c22

(b) Lines are parallel, if a1a2 = b1b2 = c1c2a1a2 = b1b2 = c1c2.

(c) Lines are perpendicular, if a1a2+b1b2+c1c2=0a1a2+b1b2+c1c2=0

4. Projection of a line on another line:

The projection of the line segment joining points P(x1, y1, z1)Px1, y1, z1 and Q(x2, y2, z2)Qx2, y2, z2 to the line having direction cosines l, m, nl, m, n is |(x2‒x1)l+(y2 ‒y1)m+(z2‒z1)n|.x2‒x1l+y2 ‒y1m+z2‒z1n.

5. Equation of a line:

(i) Unsymmetrical form of equation of a line:

If a1x+b1y+c1z+d1=0a1x+b1y+c1z+d1=0 and a2x+b2y+c2z+d2=0a2x+b2y+c2z+d2=0 are equations of two non-parallel planes, then the equation of the line of intersection is, a1x+b1y+c1z+d1=0=a2x+b2y+c2z+d2a1x+b1y+c1z+d1=0=a2x+b2y+c2z+d2.

(ii) Equation of a line in Cartesian form:

The equations of a line passing through a point (x1, y1, z1)x1, y1, z1 and having direction cosines (or direction ratios) l, m, nl, m, n are given by, x−x1l=y−y1m=z−z1nx-x1l=y-y1m=z-z1n.

(iii) Equation of a line in vector form:

The vector equation of a line passing through a point having position vector a→a→ and parallel to vector b→b→ is r→=a→+λ b→r→=a→+λ b→, where λλ is a parameter.

(iv) The equations of a line passing through points (x1, y1, z1)x1, y1, z1 and (x2, y2, z2)x2, y2, z2 are given by x−x1x2−x1=y−y1y2−y1=z−z1z2−z1x-x1x2-x1=y-y1y2-y1=z-z1z2-z1

(v) The vector equation of a line passing through points having position vectors a→a→ and b→b→ is r→=a→+λ(b→−a→)r→=a→+λ(b→-a→), where λλ is a parameter.

NCERT Solutions for Class 12 Maths for All Chapters

Other than chapter 8, here are the direct links to the articles that consists of all the details of all chapters of class 12. Students can refer to them from the links given below:

1st Chapter – Relations and Functions
2nd Chapter – Inverse Trigonometric Functions
3rd Chapter – Matrices
4th Chapter – Determinants
5th Chapter – Continuity and Differentiability
6th Chapter – Application of Derivatives
7th Chapter – Integrals
8th Chapter – Application of Integrals
9th Chapter – Differential Equations
10th Chapter – Vector Algebra
11th Chapter – Three Dimensional Geometry
12th Chapter – Linear Programming
13th Chapter – Probability

FAQs on NCERT Solutions for Class 12 Maths Miscellaneous Exercise

Q.1: What are the benefits of studying NCERT Solutions for Class 12 Maths Chapter 11 Miscellaneous Exercise?
Ans: There are several benefits of studying from Embibe’s NCERT solution. Students are well prepared for the exam and they can understand complex concepts in a simpler way which is easy to understand. Before moving on with the reference books to refer to, be thorough with NCERT solutions for class 12.

Q.2: Where can I find NCERT solutions for miscellaneous exercise chapter 11 class 12 for?
Ans: Students can find NCERT solutions here at Embibe. The solutions are designed by subject-matter experts and students can download them for and study them offline. Also, in the official website, students can find the necessary details regarding NCERT exercises and solutions.

Q.3: What are some of the topics discussed in miscellaneous chapter 11 class 12?
Ans: Some of the topics that are discussed in the miscellaneous chapter 11 class 12 for Maths are Direction Cosines and Direction Ratios of a Line, Equation of a Line in Space, Angle between Two Lines, Plane, Coplanarity of Two Lines and so on.

Q.4: Will NCERT Solution for Miscellaneous Exercise Chapter 11 Class 12 Maths improve my grades?
Ans: Yes, the NCERT solutions are reliable and are aligned with latest CBSE guidelines. Embibe NCERT solutions are comprehensive and do not leave out any questions.

Q.5: Is it important to refer to NCERT Solution for Ex 11 miscellaneous class 12?
Ans: Yes, it is important to refer to Embibe’s NCERT Solution because it helps students to understand
complex concepts and ace their exams.