Answer:
0.592 m
Explanation:
Let i and j be the unit vector in the direction of x and y respectively. And let [tex]x_A[/tex] be the x component of vector A. Therefore:
[tex]\vec{A} = x_A\hat{i}[/tex]
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 [tex]y_B[/tex] be the y component of vector B. We have
[tex]\vec{B} = -x_A\hat{i} + y_B\hat{j}[/tex]
So [tex]\vec{A} + \vec{B} = y_B\hat{j}[/tex]
This resulting vector has a magnitude 4.1 times vector A's, or [tex]4.1x_A[/tex]
[tex]y_B = 4.1x_A[/tex]
Substitute this into vector B equation we have
[tex]\vec{B} = -x_A\hat{i} + 4.1x_A\hat{j}[/tex]
Since the magnitude of vector B is 2.5
[tex]|\vec{B}| = \sqrt{x_A^2 + (4.1x_A)^2} = 2.5[/tex]
[tex]17.81x_A^2 = 2.5^2 = 6.25[/tex]
[tex]x_A^2 = 6.25 / 17.81 = 0.35[/tex]
[tex]x_A = \sqrt{0.35} = 0.592m[/tex]
So the magnitude of vector A is 0.592 m
