Vector Components

Here, we are going to discuss about the components of vector, generally vector has both the direction and the magnitude. Components, we mean as the parts of vectors, otherwise we can tell the two perpendiculars that added together means the components of the vector will give the original vector .The concept of the component vector is also the same as the rule of the vector addition .Vectors are denoted by the arrow look like as the below`->`

Rules of Vector Components

Here, we are following some rules in the components of the vectors those are given below:

• Components should always be perpendicular sometimes we called, the orthogonal components.
• The component s of the vector may be in any axis (x and the y axis) we called the horizontal or the vertical dimension.
• The direction of the components is look like the head to tail, so that we can add that vector. (because it is the components)
• So the components of the vectors are mutually independent .It will be shown through the following figure.
• If we are adding those x and y vectors we can get the resultant vector.

Examples on components of vectors

Example 1: Calculate the components of vectors along with the coordination of the position vector of P(− 4, 3)

Solution: Here, the position vector is given below `vec(OP)` = −7`veci` + 13`vecj` ,So the Component of the position vector `vec(OP)`  along x-axis is given as  − 7`veci ` . the component of  the position vector `vec(OP)`along x-axis is the  vector with the  magnitude of the 7 and here, we have to mention the direction , its direction is along the negative direction of x-axis ,because -7 is given.

So, the Component of the position vector  `vecOP` along  the y-axis is 13. The component of the position vector which as `vec(OP)` along y –axis is a vector with the  magnitude  of 13,here the direction of the vector have the positive. So,it will be the positive direction along the y-axis.

Example 2: Calculate the x-component of  d2  using the sin function the opposite value is given as 35 degree and the hypotenuse value is 33m?

Solution : Sin `theta` = opposite/hypotenuse

Sin 35 = d2x/23m

d2x = (33m)(sin 35)

d2x = (33m)(0.5736)

d2x = 18.9288m

We can check that the value, whether it is positive or negative directions. In this problem This x-component points to the right so, it is positive.

d2x = 18.928756m is the answer.