Question

A vector which has a magnitude of 8 is added to the vector P which lies along x-axis. The resultant of two vectors lies along y-axis and has magnitude twice that of P. The magnitude of P is​

Answer 1

Answer:

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}|∣BY∣=∣OA∣

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}(BY)

2

=(OY)

2

+(OB)

2

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

2

=(2B)

2

+B

2

64=3 B^{2}64=3B

2

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

2

=

3

64

Taking square root on both sides,

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

3

8