Show that the three points whose position vectors are 7j+ 10k,-i + 6j+6k and - 4i + +9j + 6k form an isosceles right angled triangle.

AB = √18 , BC=√18 and CA =4

AB²+BC²  = CA² and AB=BC

ΔABC isosceles right  angled triangle.

Step-by-step explanation:

Given vectors are  7j+ 10k,-i + 6j+6k and - 4i + +9j + 6k

A( 0,7,10), B( -1,6,6) C(-4,9,6)

AB⁻ = OB-OA = -I+6j+6k-(7j+10k) = -I-j-4k

AB =

BC = OC-OB = -4i+9j+6k-(-I+6j+6k) = -3i+3j

BC=

CA = OA-OC = 7j+10k - (- 4i + +9j + 6k ) = 4i-2j+4k

CA =

Since AB²+BC²  = CA²

And AB=BC

Therefore it follows that ΔABC is a right angled isosceles triangle



Now,

<2 = <3 [ vertically opposite angle]

Then,

<2 = <8 [corresponding angles]

Again,

<2 + <5 = 180° [straight angle]

◆ True A || B

I hope you understand...

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