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



ik how do do this just give me a few minutes and i will give it to

1)is not similar since corresponding sides are not in the same ratio

2)are similar since corresponding sides are in the same ratio and corresponding angles are equal

hope it will

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