You know that three points A,B,CA,B,C (two vectors A⃗ BA→B, A⃗ CA→C) form a plane. If you want to show the fourth one DD is on the same plane, you have to show that it forms, with any of the other point already belonging to the plane, a vector belonging to the plane (for instance A⃗ DA→D).
Since the cross product of two vectors is normal to the plane formed by the two vectors (A⃗ B×A⃗ CA→B×A→C is normal to the plane ABCABC), you just have to prove your last vector A⃗ DA→D is normal to this cross product, hence the triple product that should be equal to 00:
A⃗ D⋅(A⃗ B×AC)=0
Answer:
The first one
Step-by-step explanation:
(x-4)(x-12) hope this helps
58/6 =9.67 ( the time she runs 1 mile)
9.67 x 26.2 = 253 min
Answer:
(a) The monthly budget for housing would range from $2,291.50-$2,749.80.
the good estimate would be $2,566.48.
(b) The monthly budget for housing would range from $1,250-$1,500.
the good estimate would be $1,400.
Step-by-step explanation:
Explanation with steps are in the following attachment
Answer:
It's either a or b
Step-by-step explanation:
