Question

Vector →A which is directed along an x axis, is to be added to the vector →B which has a magnitude of 2.5m. The sum is third vector directing along the y axis with a magnitude that is 4.10times that of a vector →A What is the magnitude of →A?

Answer 1

0.592 m

Step-by-step explanation:

Let i and j be the unit vector in the direction of x and y respectively. And let
`x_A` be the x component of vector A. Therefore:

`\vec{A} = x_A\hat{i}`

Since the resulting vector only directs along the y axis, the x component of vector B must have canceled the x component of vector A. Let
`y_B` be the y component of vector B. We have

`\vec{B} = -x_A\hat{i} + y_B\hat{j}`

So
`\vec{A} + \vec{B} = y_B\hat{j}`

This resulting vector has a magnitude 4.1 times vector A's, or
`4.1x_A`

`y_B = 4.1x_A`

Substitute this into vector B equation we have

`\vec{B} = -x_A\hat{i} + 4.1x_A\hat{j}`

Since the magnitude of vector B is 2.5

`|\vec{B}| = √(x_A^2 + (4.1x_A)^2) = 2.5`

`17.81x_A^2 = 2.5^2 = 6.25`

`x_A^2 = 6.25 / 17.81 = 0.35`

`x_A = √(0.35) = 0.592m`

So the magnitude of vector A is 0.592 m