# 003 Components of a 3D force with given distances

**Problem 003**

Which of the following correctly defines the 500 N force that passes from A(4, 0, 3) to B(0, 6, 0)?

A. 256**i** - 384**j** + 192**k** N

B. -256**i** + 384**j** - 192**k** N

C. -384**i** + 192**j** - 256**k** N

D. 384**i** - 192**j** + 256**k** N

**Solution 003**

${\bf r}_{AB} = -4{\bf i} + 6{\bf j} - 3{\bf k} \, \text{ m}$

Unit vector from A to B:

${\bf \lambda}_{AB} = \dfrac{{\bf r}_{AB}}{r_{AB}}$

${\bf \lambda}_{AB} = \dfrac{-4{\bf i} + 6{\bf j} - 3{\bf k}}{\sqrt{(-4)^2 + 6^2 + (-3)^2}}$

${\bf \lambda}_{AB} = -0.5121{\bf i} + 0.7682{\bf j} - 0.3841{\bf k}$

Rectangular representation of **F**:

${\bf F} = F \, {\bf \lambda}_{AB}$

${\bf F} = 500(-0.5121{\bf i} + 0.7682{\bf j} - 0.3841{\bf k})$

${\bf F} = -256{\bf i} + 384{\bf j} - 192{\bf k} \, \text{ N}$

Answer: **B**

**Calculator Technique**

Use VECTOR mode: [MODE] → 8:VECTOR

This mode is made primarily for vector quantities, thus, handling forces in 3D is straightforward.

Enter position vector:

[MODE] → 8:VECTOR → 1:VctA → 1:3**r** = VctA = [ -4 6 -3 ]

Solve for vector **F**: AC**F** = 500[×]([SHIFT] → [5 VECTOR] → 3:VctA ÷ [SHIFT] → [hyp Abs]( → [SHIFT] → [5 VECTOR] → 3:VctA) ) =

Calculator display: 500×(VctA÷Abs(VctA))

**F** = [ -256 384 -192 ] *answer*

Note: the unit vector λ is:

λ = [SHIFT] → [5 VECTOR] → 3:VctA ÷ [SHIFT] → [hyp Abs]( → [SHIFT] → [5 VECTOR] → 3:VctA)