Question

The vertices A B C of a triangle are (2,-1,-3) (4,2,3) and (6,3,4) respectively. Show that vector AB=(2,3,6) and AC=9.​

Answer 1

Try this option:

1) vector AB=(2;3;6).

$(X_B-X_A;Y_B-Y_A;Z_B-Z_A); \ => (4-2;2+1;3+3); \ =>(2;3;6).$

2) |AC|=9.

$|AC|=\sqrt{(X_C-X_A)^2+(Y_C-Y_A)^2+(Z_C-Z_A)^2}=\sqrt{(6-2)^2+(3+1)^2+(4+3)^2}=\sqrt{81}=9.$