Question

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

Answer 1

Answer:

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 = $\sqrt{1+1+16} = \sqrt{18}$

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

BC=$\sqrt{9+9} =\sqrt{18}$

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

CA = $\sqrt{16+4+16} =\sqrt{36} =4$

Since AB²+BC²  = CA²

And AB=BC

Therefore it follows that ΔABC is a right angled isosceles triangle