Please help me with this physics question. I think I am making this harder than it seems.?

Question

Find the magnitude of the vector sum r of the vector displacements c and d whose components in meters along three perpendicular directions are cx= 7.0, cy= -3.5, cz= -6.4; dx= 4.1, dy= -2.4, dz= 3.3;

Answer 1

For this problem, you have to use the vector addition rule and the vector magnitude formula.

r = C+D = <C1+D1, C2+D2, C3+D3>

= <7.0, -3.5, -6.4> + <4.1, -2.4, 3.3>

= <(7.0 + 4.1), (-3.5 - 2.4), (-6.4 + 3.3)>

= <11.1, -5.9, -3.1>

||r|| = (ri² + rj² + rk²)^(1/2)

= ((11.1)² + (-5.9)² + (-3.1)²)^(1/2)

= 12.9

Answer 2

Add the x directions to get total x, add they y directions to get total y, and add the z directions to get total z.  Then take those three numbers, square them, add them together, and square-root that sum:

magnitude: sqrt[ (cx+dx)^2 + (cy+dy)^2 + (cz+dz)^2)

Your answer is: 12.9