# Supplementary Exercise Q10

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Supplementary Exercise Q10 If the vectors 𝐴 = a 𝑖 + 𝑗 + 𝑘, 𝐵 = 𝑖 + b 𝑗 + 𝑘 and 𝐶 = 𝑖 + 𝑗 + c 𝑘 are coplanar, then 11−𝑎 + 11−𝑏 + 11−𝑐 = 1, where a, b, c ≠ 1 Since Vectors 𝐴, 𝐵, 𝐶 are coplanar , ∴ 𝐴 𝐵 𝐶 = 0 𝐴 𝐵 𝐶 = 𝑎111𝑏111𝑐 = 0 C1 C1 C3 C2 C2 C3 𝑎−1010𝑏−111−𝑐1−𝑐𝑐 = 0 𝑎−1010𝑏−111−𝑐1−𝑐𝑐 = 0 Expanding Determinant along C3 1 0(1 − c) − (1 − c)(b − 1) −1 (a − 1) (1 − c) − 0 + c (a − 1) (b − 1) − 0 (1 − c) (1 − b) − (a − 1) (1 − c) + c (1− a) (1 − b) = 0 (1 − c) (1 − b) + (1 − c) (1 − a) + c (1 − a) (1− b) = 0 Dividing L.H.S and R.H.S by (1 − a) (1 − b) (1 − c) 1 − 𝑐 1 − 𝑏 + 1 − 𝑐 1 − 𝑎 + 𝑐 1 − 𝑎 1 − 𝑏 1 − 𝑎 1 − 𝑏 1 −𝑐 =0 11 − 𝑎+ 11 − 𝑏+ 𝑐1 − 𝑐=0 11 − 𝑎+ 11 − 𝑏+ 𝑐1 − 𝑐=0 Adding 1 both sides 11 − 𝑎+ 11 − 𝑏+ 𝑐1 − 𝑐+1=1 11 − 𝑎+ 11−𝑏+ 𝑐 + 1 − 𝑐1 − 𝑐=1 11 − 𝑎+ 11 − 𝑏+ 11 − 𝑐=1 Hence Proved

Supplementary Example 1

Supplementary Example 2

Supplementary Example 3

Supplementary Example 4

Supplementary Example 5

Supplementary Example 6

Supplementary Exercise Q1

Supplementary Exercise Q2

Supplementary Exercise Q3

Supplementary Exercise Q4

Supplementary Exercise Q5

Supplementary Exercise Q6

Supplementary Exercise Q7

Supplementary Exercise Q8

Supplementary Exercise Q9

Supplementary Exercise Q10 You are here

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.