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



here you hope this .

step-by-step explanation:

hypotenuse = √(15² + 8²)

=  √(225 +64)

=  √(289) = 17

sin(x) = cos(y) = 8/17

they are identical so long as you are doing the sine of one angle and the cosine of the only other angle that isn't the right angle.

step-by-step explanation:

the options are kind of confusing. i can tell you what i see going on here when i look at the graph.

the class and exam medians are closely related, and it looks like the low outlier on the exam did pull the median lower. [tex][/tex]