Last updated at March 11, 2017 by Teachoo

Transcript

Ex 10.2, 17 Show that the points A, B and C with position vectors, = 3 4 4 , = 2 + and = 3 5 , respectively form the vertices of a right angled triangle. Position vectors of vertices A, B, C of triangle ABC are = 3 4 4 , = 2 1 + 1 = 1 3 5 We know that two vectors are perpendicular to each other, i.e. have an angle of 90 between them, if their scalar product is zero. So, if CA . AB = 0, then CA AB & CAB = 90 Now, AB = = (2 i 1 j + 1 k ) (3 i 4 j 4 k ) = (2 3) i + ( 1 + 4) j + (1 + 4) k = 1 i + 3 j + 5 k BC = = (1 i 3 j 5 k ) (2 i 1 j + 1 k ) = (1 2) i + ( 3 + 1) j + ( 5 1) k = -1 i 2 j 6 k CA = = (3 i 4 j 4 k ) (1 i 3 j 5 k ) = (3 1) i + ( 4 + 3) j + ( 4 + 5) k = 2 i 1 j + 1 k Now, . = ( 1 i + 3 j + 5 k ) . (2 i 1 j + 1 k ) = ( 1 2) + (3 1) + (5 1) = ( 2) + ( 3) + 5 = 5 + 5 = 0 So, AB . CA = 0 Thus, AB and CA are perpendicular to each other. Hence, ABC is a right angled triangle.

Ex 10.2

Ex 10.2, 1

Ex 10.2, 2

Ex 10.2, 3

Ex 10.2, 4 Important

Ex 10.2, 5 Important

Ex 10.2, 6

Ex 10.2, 7 Important

Ex 10.2, 8

Ex 10.2, 9

Ex 10.2, 10 Important

Ex 10.2, 11

Ex 10.2, 12

Ex 10.2, 13 Important

Ex 10.2, 14

Ex 10.2, 15 Important

Ex 10.2, 16

Ex 10.2, 17 Important You are here

Ex 10.2, 18 Important

Ex 10.2, 19 Important

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.