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 = [tex]\sqrt{1+1+16} = \sqrt{18}[/tex]
BC = OC-OB = -4i+9j+6k-(-I+6j+6k) = -3i+3j
BC=[tex]\sqrt{9+9} =\sqrt{18}[/tex]
CA = OA-OC = 7j+10k - (- 4i + +9j + 6k ) = 4i-2j+4k
CA = [tex]\sqrt{16+4+16} =\sqrt{36} =4[/tex]
Since AB²+BC² = CA²
And AB=BC
Therefore it follows that ΔABC is a right angled isosceles triangle
