In Triangle ABC AB = 17, AC = 15, BC = 8 cm find the product of vectors AB * AC, BA * BC, CA * CB

In the triangle ABC AB ^ 2 = 289, BC ^ 2 + AC ^ 2 = 64 + 225 = 289.

Since AB ^ 2 = BC ^ 2 + AC ^ 2, then triangle ABC is rectangular, angle C = 90.

Let us determine the cosines of the angles ABC and BAC.

CosABC = BC / AB = 8/17.

CosBAC = AC / AB = 15/17.

Then the product of vectors is equal to:

↑ AB * ↑ AC = | AB | * | AC | * CosBAC = 17 * 15 * (15/17) = 225.

↑ BA * ↑ BC = | BA | * | Sun | * CosABC = 17 * 8 * (8/17) = 64.

↑ CA * ↑ СB = | CA | * | CB | * Cos90 = 15 * 8 * 0 = 0.

Answer: ↑ AB * ↑ AC = 225, ↑ BA * ↑ BC = 64, ↑ CA * ↑ СB = 0.