Given a vector with components, x = 2 , y = 5 , find the vector’s magnitude and direction.?

Question

Given a vector with components, x = 2 , y = 5 , find the vector’s magnitude and direction.

please explain

Answer 1

think of this as a right angle triangle... you have the base which is x and your height which is y now you have to figure out the 3rd leg so how do you do it? Well, my friend lol, you use xsquared + ysquared = 3rd leg squared that is
2squared + 5squared = 3rd leg squared
4 + 25 = 3rd leg squared
29 = 3rd leg squared
square root of 29 = 3rd leg
3rd leg ='s 5.385.
This is your magnitute... now to find the direction again think of it as a right triangle... you are trying to figure out the angle between the 3rd leg and the base. Now we know the value of 3rd leg (magnitude) and we know te value of the base (2) so what trig function will give us the angle? It is COS so the angle is cos(z) = 2/5.385 angle z = inverse cos(2/5.385)
z = 68.197 degrees

And now we know that this is in the first quadrant on the graph. And usually North is up and East is to the right of north. So the direction of this vector is 68.197North East

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Magnitude = 5.385
Direction = 68.197NorthEast

Answer 2

your vector is 2i + 5j

Mathematically, magnitude is always given by sqrt(x^2 + y^2)
Substituting for x=2 and y=5

If my understanding of what you need in terms of vector direction is correct, it can be expressed as an angle, so trigonometry may be adequate. (inverse tan function)

Answer 3

V= squareroot of 4+25= squareroot of 29 units is the magnitude. and the direction is arc tan of squareroot of 29. then u know if less 90= Quad one. etc./..

Answer 4

well its quite simple... the magnitude is (x^2 + y^2)^0.5
ie square root of 2 square + 5 square = (4+25)^0.5

so magnitude = square root of 29.

and the direction is given by inverse tan of y/x i.e atan(5/2).