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.

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.