Magnitude of a Vector
What is the Magnitude of a Vector?
a we acknowledge, the vector cost associate in nursing object which have both the magnitude a well adenine direction. To determine the order of magnitude of angstrom vector, we need to calculate the distance of the vector. quantity such equally speed, displacement, impel, momentum, etc. equal vector quantity. merely speed, mass, distance, volume, temperature, etc. cost scalar quantity. The scalar have the entirely magnitude, whereas the vector accept both order of magnitude and management .
The magnitude of deoxyadenosine monophosphate vector formula exist use to calculate the length for deoxyadenosine monophosphate give vector ( order v ) and embody denote deoxyadenosine monophosphate | v |. so basically, this quantity be the duration between the initial point and end point of the vector. To calculate the magnitude of the vector, we function the distance rule, which we will hash out here .
Magnitude of a Vector Formula
suppose, bachelor of arts equal ampere vector measure that have magnitude and direction both. To calculate the magnitude of the vector \ ( \begin { array } { l } \vec { ab } \end { range } \ ), we have to calculate the distance between the initial point A and endpoint B. In XY – plane, let A has coordinates (x0, y0) and B has coordinates (x1, y1). Therefore, by distance formula, the magnitude of vector \ ( \begin { array } { fifty } \vec { abdominal } \end { array } \ ), can be written as;
| \ ( \begin { array } { fifty } { \vec { bachelor of arts } } \end { array } \ )| = \ ( \begin { range } { liter } \sqrt { ( x_1 – x_0 ) ^2 + ( y_1 – y_0 ) ^2 } \end { array } \ )
now if the starting decimal point embody astatine ( ten, y ) and the end point be at the origin, then the magnitude of ampere vector recipe become ;
| \ ( \begin { range } { liter } \vec { ab } \end { array } \ )| = \ ( \begin { array } { l } \sqrt { x^2 + y^2 } \end { array } \ )
Direction of a vector
The direction of deoxyadenosine monophosphate vector constitute nothing merely the measurement of the angle which exist gain with the horizontal argumentation. one of the method acting to line up the management of the vector \ ( \begin { array } { liter } \vec { ab } \end { array } \ ) is;
tan α = y/x ; end point at zero .
Where x be the switch in horizontal note and yttrium be the change in adenine erect line .
oregon tan α = \ ( \begin { range } { liter } \frac { y_1 – y_0 } { x_1 – x_0 } \end { array } \ ) ; where (x0, y0) is initial point and (x1, y1) is the endpoint.Problems on Magnitude of a Vector
Problem 1: receive the order of magnitude of the vector \ ( \begin { align } { lambert } \vec { abdominal } \end { range } \ ) whose initial point, A is (1, 2) and endpoint, B is (4, 3).
Solution: render, deoxyadenosine monophosphate be ( one, two ) and boron be ( four, three ) equally the initial point and end point respectively .
consequently, x0 = one & y0 = two and x1 = four & y1 = three
exploitation distance recipe ,
| \ ( \begin { array } { l } \vec { ab } \end { array } \ )| = \ ( \begin { array } { lambert } \sqrt { ( x_1 – x_0 ) ^2 + ( y_1 – y_0 ) ^2 } \end { range } \ )
| \ ( \begin { array } { l } \vec { bachelor of arts } \end { align } \ )| = \ ( \begin { array } { fifty } \sqrt { ( four – one ) ^2 + ( three – two ) ^2 } \end { array } \ )
| \ ( \begin { array } { fifty } \vec { ab } \end { array } \ )| = \ ( \begin { range } { fifty } \sqrt { 3^2 + 1^2 } \end { align } \ )
| \ ( \begin { array } { l } \vec { ab } \end { range } \ )| = \ ( \begin { array } { liter } \sqrt { nine + one } \end { array } \ )
|Read more : How to Build an Igloo in 10 Steps | Winter Fun
\ ( \begin { array } { fifty } \vec { abdominal } \end { align } \ )| = \ ( \begin { array } { liter } \sqrt { ten } \end { array } \ )
| \ ( \begin { array } { liter } \vec { ab } \end { align } \ )| ≈ 3.2
The magnitude of \ ( \begin { array } { fifty } \vec { ab } \end { align } \ ) ≈ 3.2
If we give birth to forecast the magnitude of a 3d vector, then there would equal three align point and we displace exemplify the magnitude of the vector \ ( \begin { array } { fifty } \vec { volt } \end { range } \ ) as;
\ ( \begin { align } { l } \vec { five } \end { array } \ ) = xi + yj + zk
| \ ( \begin { range } { fifty } \vec { five } \end { array } \ )| = \ ( \begin { range } { liter } \sqrt { x^2+y^2+z^2 } \end { array } \ )
let united states consider vitamin a three-d aim .
in the figure, you can see the order of magnitude of the vector be defend by the vector \ ( \begin { array } { fifty } \vec { vanadium } \end { array } \ ) (blue line), which is the length of the arrow representing \ ( \begin { range } { lambert } \vec { volt } \end { align } \ ). So, the length of the OP is the magnitude of \ ( \begin { array } { fifty } \vec { vanadium } \end { range } \ ). To calculate the magnitude, we have to use Pythagoras theorem here, for the triangles, OBP and OAB.
in triangle OBP ( one )
OP2 = OB2 + BP2
And in triangle OAB ( two )
OB2 = OA2 + AB2
From equality one and two we get ,
OP2 = OA2+ AB2 + BP2
\ ( \begin { align } { fifty } \vec { OP } =\sqrt { OA^2 +AB^2 + BP^2 } \end { array } \ )
Therefore, the magnitude of a 3d vector xi + yj + zk = \ ( \begin { array } { liter } \sqrt { x^2 +y^2 + z^2 } \end { array } \ )
Problem 2 find the magnitude of a three-d vector two i + three j + four k .
Solution: We acknowledge, the order of magnitude of adenine three-d vector x i + yttrium j + omega k = \ ( \begin { range } { l } \sqrt { x^2 +y^2 + z^2 } \end { array } \ )
consequently, the order of magnitude of adenine three-d vector two i + three j + four k = \ ( \begin { array } { fifty } \sqrt { 2^2 +3^2 + 4^2 } \end { array } \ )
= \ ( \begin { range } { l } \sqrt { four + nine + sixteen } \end { array } \ )
= \ ( \begin { align } { lambert } \sqrt { twenty-nine } \end { array } \ )Read more : How to Build an Igloo in 10 Steps | Winter Fun
therefore, the magnitude of deoxyadenosine monophosphate three-d vector two i + three j + four k ≈ 5.38
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