Question

A vector A which has magnitude 8.0 is added to a vector B which lies on x-axis.The sum of these two vectors lies on y-axis and has a magnitude twice of the magnitude of vector B.The magnitude of vector B is

Answer 1

Let unit vector along x and y axes by i and j respectively

According to question
8 + Bi = 2Bj
8 = 2Bj - Bi
8 = sqrt((2B)^2 + B^2)
8 = sqrt(5B^2)
8 = Bsqrt(5)
B = 8 / sqrt(5)

Answer 2

The magnitude of vector B is $\bold{\frac{8}{\sqrt{3}}}$.

Solution:

Let us consider O as the origin point and OB as the vector B on x axis and OA as another vector with magnitude 8. Let us consider the sum of OA and OB in y axis as OY vector whose magnitude is twice the magnitude of OB.

So if we sum up the data’s we have, then we get that  

OB in x axis = B

OA = 8

OY = 2B

Draw a vector connecting Y with B named as BY whose magnitude will be same as vector OA

So $|\mathrm{BY}|=|\mathrm{O} \mathrm{A}|$

In this case, with the datas provided it can noted that a right angled triangle BOY has been formed whose sides OY, BY is known to us and we need to determine the third side OB.

So using Pythagoras theorem,

$(B Y)^{2}=(O Y)^{2}+(O B)^{2}$

$(8)^{2}=(2 B)^{2}+B^{2}$

$64=3 B^{2}$

$B^{2}=\frac{64}{3}$

Taking square root on both sides,

$\bold{B=\frac{8}{\sqrt{3}}}$