NCERT solution Class 12 three dimensional geometry

NCERT Solutions

Class 12 Maths

Three Dimensional Geometry

NCERT Questions

Ex.11.1 Q.1
Ex.11.1 Q.2
Ex.11.1 Q.3
Ex.11.1 Q.4
Ex.11.1 Q.5
Ex.11.2 Q.1
Ex.11.2 Q.2
Ex.11.2 Q.3
Ex.11.2 Q.4
Ex.11.2 Q.5
Ex.11.2 Q.6
Ex.11.2 Q.7
Ex.11.2 Q.8
Ex.11.2 Q.9
Ex.11.2 Q.11
Ex.11.2 Q.11
Ex.11.2 Q.12
Ex.11.2 Q.13
Ex.11.2 Q.14
Ex.11.2 Q.15
Ex.11.2 Q.16
Ex.11.2 Q.17
Ex.11.3 Q.1
Ex.11.3 Q.2
Ex.11.3 Q.3
Ex.11.3 Q.4
Ex.11.3 Q.5
Ex.11.3 Q.6
Ex.11.3 Q.7
Ex.11.3 Q.8
Ex.11.3 Q.9
Ex.11.3 Q.10
Ex.11.3 Q.11
Ex.11.3 Q.12
Ex.11.3 Q.13
Ex.11.3 Q.14
Ex.Misc. Q.1
Ex.Misc.Q.2
Ex.Misc.Q.3
Ex.Misc.Q.4
Ex.Misc.Q.5
Ex.Misc.Q.6
Ex.Misc.Q.7
Ex.Misc.Q.8
Ex.Misc.Q.9
Ex.Misc.Q.10
Ex.Misc.Q.11
Ex.Misc.Q.12
Ex.Misc.Q13
Ex.Misc.Q.14
Ex.Misc.Q.15
Ex.Misc.Q.16
Ex.Misc.Q.17
Ex.Misc.Q.18
Ex.Misc.Q.19
Ex.Misc.Q.20
Ex.Misc.Q.21
Ex.Misc.Q.22
Ex.Misc.Q.23

Ex.11.3 Q.5

Find the vector and Cartesian equation of the planes

(a) that passes through the point (1, 0, −2) and the normal to the plane is i + j - k.

(b) that passes through the point (1, 4, 6) and the normal vector to the plane is i – 2j + k.

(a) The position vector of point (1, 0, −2) is a = i – 2k

The normal vector N perpendicular to the plane is N = i + j - k

The vector equation of the plane is given by,

(r - a).N = 0

=> [r – (i – 2k)].(i + j - k) = 0 …………1

r is the position vector of any point P(x, y, z) in the plane.

So, r = xi + yj + zk

Therefore, equation 1 becomes

[(xi + yj + zk) – (i – 2k)].(i + j - k) = 0

=> [(x – 1)i + yj + (z + 2)k] .(i + j - k) = 0

=> (x – 1) + y - (z + 2) = 0

=> x + y – z – 3 = 0

=> x + y – z = 3

This is the Cartesian equation of the required plane.

(b) The position vector of the point (1, 4, 6) is a = i + 4j + 6k

The normal vector perpendicular to the plane is N = i – 2j + k

The vector equation of the plane is given by,

(r - a).N = 0

=> [r – (i + 4j + 6k)].(i - 2j + k) = 0 …………1

r is the position vector of any point P(x, y, z) in the plane.

So, r = xi + yj + zk

Therefore, equation 1 becomes

[(xi + yj + zk) – (i + 4j + 6k)].(i - 2j + k) = 0

=> [(x – 1)i + (y - 4)j + (z - 6)k] .(i - 2j + k) = 0

=> (x – 1) – 2(y - 4) + (z - 6) = 0

=> x - 2y + z + 1 = 0

This is the Cartesian equation of the required plane.

Three Dimensional Geometry Class 12 Maths NCERT Chapter 11 NCERT Solutions

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NCERT solutions of Three Dimensional Geometry Class 12 Maths - all questions of all exercises are solved. Q. No. 1, 2, 3, and so on. Complete explanation and detailed solution makes them the best videos for Class 12 Maths Three Dimensional Geometry NCERT solutions.