# Tips for solving word problems of vector?

I will be taking PreCalc next school year. I have prepared for it by using Khan Academy. Now, I am learning vectors. I can calculate the sum and difference of vectors. However, I am not good at solving word problems of vectors. Could you give me some tips for solving it well?

### 1 Answer

- SeanLv 51 month ago
An example would be helpful oh well let's make something up

Two friends A and B go camping. One morning A goes out for a walk with speed of 4K/Hr while B whose turn it is does the dishes. A who has bad handwriting leaves a message saying he will be travelling north for an hour and then west for three quarters of an hour and invites B to meet him there. B mixes up the order directions and times and sets out fifteen minutes later not realizing that his walking speed is only 3K/Hr after 75 minutes he realizes his error and calls A on a cell phone. A advises B to stop and said that he would join him where B is

i) where is A when B calls relative to the campsite

ii)where is B when the call is made relative to the campsite

iii) how far must A travel and in what direction

i) At the time of call one and one half hours (75 minutes and the quarter hour head start) have passed for A who has gone one hour north and only a half hour west

this is a right triangle 1*4 kilometres and 1/2* 4 kilometres west

changing to vector form with campsite at (0,0) the position is (-2,4) so

cos(alpha) =-2/sqrt((-2)^2 + 4^2)) = -2/sqrt(20) = -1/sqrt(5) = 116.6 degrees from the east relative to the campsite and sqrt(20) kilometres 4.5 kilometres away

ii) B has gone one hour west and one quarter hour north at 3K/hr

so the position is (-3*1,3/4) so cos(beta) = -3/sqrt((-3)^2 + (3/4)^))

166.0 degrees from the campsite measuring from due east a distance of

sqrt(9 + 9/16) = sqrt(153/16) = 3.09 kilometres

iii) A has reached the position (-2,4) and B (-3,3/4) to get the direction A must travel subtract A's vector (from which he sets out) from B's vector

(-3 - -2,4-3/4) = (-1,-13/4) which is cos(gamma) = -1/sqrt((-1)^2 +(13/4)^2)) = -1/sqrt(185/16) for a distance of 3.4 kilometres at a bearing of 252.9 degrees from due east

NOTES for practical problems converting vectors into an (a,b) form is usually the easiest way to proceed for example a bearing southeast for 20 kilometres is

20(cos theta, sin theta) since southeast is the fourth quadrant gives 20 (cos (305-270), sin(305-270))

assuming angles are measured counterclockwise the formula is

0 <= theta <90 convert theta to theta - 0

180 >theta>= 90 theta converts to theta -90

270> theta >=180 theta converts to theta - 180

360 > theta >=270 theta converts to theta - 270

if coordinates (a,b) are turned into k(cos(x),sin(x))

calculate

"cos" as |a|/sqrt(a^2 + b^2)

"sin" as |b|/sqrt(a^2 + b^2)

"tan" as |b|/|a|

a>=0,b>=0 arccos("cos"),arcsin("sin"),arctan("tan")

a < 0 b>= 0 180-arccos("cos"),180-arcsin("sin"),180-arctan("tan")

a<0 b<=0 180 + arccos("cos"), 180 + arcsin("sin"), 180 + arctan("tan")

a>0,b<=0 360 - arccos("cos"),360-arcsin("sin"),360-arctan("tan")