Question

A vector v has initial point (22, 25) and terminal point ( -11, 13). Find its magnitude and direction

Answer 1

v = <-33,12>
||v|| = sqrt [ 33^2 + 12^2 ]

which is sqrt 265 yes... but is that the answer??

Answer 2

@surjithayer ^

Answer 3

what about the direction?

Answer 4

vector v= (-11-22)i+(13-25)j=-33i-12j
\[magnitude=\left| v \right|=\left| \sqrt{\left( -33 \right)^{2}+\left( -12 \right)^{2}} \right|\]
\[=\sqrt{1089+144}=\sqrt{1233}=3\sqrt{137}\]
\[unit~ normal=\frac{ v }{ \left| v \right| }=\frac{ -33i-12j }{ 3\sqrt{137} }=\frac{-11i-4j }{ \sqrt{137} }\]

Answer 5

thats the unit vector in direction of v.... but its not asking for a unit vector.. just a regular vector @surjithayer

Answer 6

correction. it is unit vector not unit normal.

Answer 7

so how do I find the direction of the vector without reducing its size down to 1 unit?

Answer 8

v is regular vector.

Answer 9

so... to answer the question..
I just give the magnitude which is 3 sqrt(137)
and v = <-33, 12> for direction?

Answer 10

the direction is the part of the answer that confuses me... This is a practice exam for studying purposes, but i really want to understand this stuff so i can ace the exam tomorow

Answer 11

v=<-33,-12>
for direction we generally give direction ratios or direction cosines.
here d.r's are <-33,-12>
d.c's are <-11/sqrt137,-4/sqrt137>

Answer 12

oh okay.. so that explains it then. Thankyou very much sir... I really appreciate it

Answer 13

yw

Answer 14

so its <a2/||v|| , b2/ ||v|| > right ??