In the parallelogram ABCD ∟A = 30 °, AB = 2√3 BC = 5. Find scalar vectors: a) AD • AB; b) BA • BC; c) AD • BH

∟А = 30 °; AB = 2√3 = CD; BC = 5 = AD.

AD * AB = | AD | * | AB | * cos 30 ° = 5 * 2√3 * √3 / 2 = 15;

∠B = (360 ° – 60 °) / 2 = 150 °;

BA * BC = | BA | * | BC | * cos 150 ° = | BA | * | BC | * sin 60 ° = 2√3 * 5 * √3 / 2 = 15;

AB * BH = | AB | * | BH | * cos 60 °;

| BH | = 1/2 | AB | (opposite an angle of 30 degrees lies a leg equal to half of the hypotenuse) = √3;

∠ABH = 90 ° – 30 ° = 60 ° (from triangle ABH, where H is right angle)

AB * BH = | AB | * | BH | * cos 60 ° = 2√3 * √3 * 1/2 = 3.