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### Course: Algebra (all content)>Unit 19

Lesson 8: Component form of vectors

# Vector components from magnitude & direction

Sal finds the components of a couple of vectors given in magnitude and direction form.

## Want to join the conversation?

• At , how did Sal come up with 5√2?
I didn't quite understand that part, can someone help me on this please?
Thanks!
• It comes from knowing the unit circle and trigonometric functions.
The cosine of 45 degrees is √2/2, therefore 10(√2/2) = 5√2.
You should familiarize yourself with the unit circle, as these types of trig questions are more frequent in calculus.
Print out this image and have it handy when doing your work. There are videos on Khan that talk about it as well.
https://lelandmath.files.wordpress.com/2013/10/screen-shot-2013-10-14-at-7-19-46-pm.png
• If x component is 4 cos50 then its y component can also be written as 4 cos(90-θ) so its gonna be 4 sin40 right?? Just a little bit doubt in it?
• In my textbook the <?,?> signs are used. What is the difference between <?,?> and just (?,?)?
• Typically, <x,y> is a vector and (x,y) is a point.
• Hi, quick question, is there a way to find the components of a vector with just the angle and nothing else? Thanks!
• No, because there are infinitely many vectors with just that angle.
• If the x and y vector components of Vector A are 11 and 7,

what is the magnitude of Vector A? Round your answer to the nearest hundredth.
• ||A||=sqrt(11^2+7^2)=sqrt(121+49)=sqrt(170)=13.05
Hope it helps..
• (1 vote)
• position vector: OA=4i+k, OB=5i-2j-2k, OC=i+j, OD=-i-4k
point: a=(4,0,1) b=(5,-2,-2) c=(1,1,0) d=(1,0,-4)
vector: AB =(5-4,-2-0,-2-1)=(1,-2,-3) DC=(1--1,1-0,0--4)=(2,1,-4)
symmetric form : line AB : (x-4)/1=(y-0)/-2=(z-1)/-3
line CD : (x-1)/2=(y-1)/1=(z-0)/4
solve for x,z (x-4)=(z-1)/-3 , (x-1)/2=z/4
-3x+13=z and 2x-2=z
x=3 z=4
apply to line formula : (3-4)/1=(y-0)/-2=(4-1)/-3 , (3-1)/2=(y-1)/1=(4-0)/4
get y/-2=-1 , y-1=1 y=2
so point (3,2,4) satisfied both line AB and CD ,therefore two line intersect.

point p=(1,5,6) line AB= (x-4)/1=(y-0)/-2=(z-1)/-3 , so point x=4+n , y=-2n , z=1-3n on line AB.
let q on AB , vector PQ =(4+n-1 , -2n-5 , 1-3n-6), if PQ perpendicular AB , then (n+3)*1+(-2n-5)*-2+(-3n-5)*-3=0 (point product of perpendicular vector=0)
n+3+4n+10+9n+15=0 , n=-2 . PQ=(4-2-1 , 4-5 , 1+6-6)=(1,-1,1) magnitude =squart(1^2+1^2+1^2)=squart(3)
• Can you please tell me what will happen if angle is 180 degrees.

Can ever component will be greater than vector itself?