NCERT solution Class 11 introduction to three dimensional geometry

NCERT Solutions

Class 11 Maths

Introduction to Three Dimensional Geometry

NCERT Questions

Ex.12.1 Q.1
Ex.12.1 Q.2
Ex.12.1 Q.3
Ex.12.1 Q.4
Ex.12.2 Q.1
Ex.12.2 Q.2
Ex.12.2 Q.3
Ex.12.2 Q.4
Ex.12.2 Q.5
Ex.12.3 Q.1
Ex.12.3 Q.2
Ex.12.3 Q.3
Ex.12.3 Q.4
Ex.12.3 Q.5
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.2

Find the lengths of the medians of the triangle with vertices A (0, 0, 6), B (0, 4, 0) and (6, 0, 0).

Let AD, BE, and CF be the medians of the given triangle ABC.

Since AD is the median, D is the mid-point of BC.

So, coordinates of point D = {(0 + 6) ÷ 2, (4 + 0) ÷ 2, (0 + 0) ÷ 2} = (3, 2, 0)

AD = √ {(0 - 3)² + (0 - 2)² + (6 - 0)²} = √ (9 + 4 + 36) = √49 = 7

Since BE is the median, E is the mid-point of AC.

So, coordinates of point E = {(0 + 6) ÷ 2, (4 + 0) ÷ 2, (6 + 0) ÷ 2} = (3, 0, 3)

BE = √ {(3 - 0)² + (0 - 4)² + (3 - 0)²} = √ (9 + 16 + 9) = √34

Since CF is the median, EF is the mid-point of AB.

So, coordinates of point F = {(0 + 0) ÷ 2, (0 + 4) ÷ 2, (6 + 0) ÷ 2} = (0, 2, 3)

Length of CF = √ {(6 - 0)² + (0 - 2)² + (0 - 3)²} = √ (36 + 4 + 9) = √49 = 7

Thus, the lengths of the medians of ∆ABC are 7, √34 and 7.

Introduction to Three Dimensional Geometry Class 11 Maths NCERT Chapter 11 NCERT Solutions

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