I have no idea how my teacher gets the answers for " Example 2.6" and "Example 2.7". Honestly, I am terrible at physics and Vector stuff.

So,please take your time to explain in detail to me how my teacher gets the answers for the questions below, using more calculation,less wording and more tabling or drawing explanations!

Example 2.6:

Example 2.6 Given: Two forces are acting on a pipe as shown in the figure. =Bl1b g - . !._f, 7 r-wn Find: The magnitude and the gt > \\ coordinate direction o it S\\A s e angles of the resultant e At T R -' force. Plan: 1) Find the forces along CA and CB in the Cartesian vector form. 2) Add the two forces to get the resultant force, F,. 3) Determine the magnitude and the coordinate angles of F,. \fExample 2.7 The frame shown in Fig. 243 is subjected to a horizontal force F = {300j] N. Determine the magnitudes of the components of this force parallel and perpendicular to member AB. B F=[300j|N s (a) (b} Fig. 2-43 SOLUTION The magnitude of the component of F along AB is equal to the dot product of F and the unit vector up, which defines the direction of AB, Fig. 2-43b. Since 2i + 6j + 3k UB 0.2861 + 0.857j + 0.429k V (2)+ (6)+ (3)2 then FAB = FOOS 0 = F . us = (300j) . (0.2861 + 0.857j + 0.429k) = (0)(0.286) + (300)(0.857) + (0)(0.429) = 257.1 N Ans.Since the result is a positive scalar, FAR has the same sense of direction as up. Fig. 2-43b. Expressing FAB in Cartesian vector form, we have FAB = FABUR = (257.1 N)(0.2861 + 0.857j + 0.429k) = {73.51 + 220j + 1 10k } N Ans. The perpendicular component, Fig. 2-43b, is therefore F =F - FAB = 300j - (73.51 + 220j + 110k) = {-73.51 + 79.6j - 110k } N Its magnitude can be determined either from this vector or by using the Pythagorean theorem, Fig. 2-43b: FI =VP - FAB = V (300 N)- - (257.1 N)- = 155 N Ans