When vector overrightarrow (B) is subtracted from vector overrightarrow (A)=4hat (i)-6hat (j) the resulting vector points in the positive x-direction and has a magnitude of 4. The magnitude of dot (B) is A. 6

Let's denote the vector $\overrightarrow{B}$ as $B_x\hat{i} + B_y\hat{j}$. When vector $\overrightarrow{ is subtracted from vector $\overrightarrow{A}$, we get:<br /><br />$\overrightarrow{A} - \overrightarrow{B} = (4\hat{i} - 6\hat{j}) - (B_x\hat{i} + B_y\hat{j})$<br /><br />This simplifies to:<br /><br />rightarrow{A \overrightarrow{B} = (4 - B_x)\hat{i} + (-6 - B_y)\hat{j}$<br /><br />According to the given information, the resulting vector points in the positive x-direction and has a magnitude of 4. Therefore, we have:<br /><br />$4 - B_x = 4$ (since the vector points in the positive x-direction)<br /><br />Solving for $B_x$, we get:<br /><br />$B_x = 0$<br /><br />And the magnitude of $\overrightarrow{B}$ is given by:<br /><br />$|\overrightarrow{B}| = \sqrt{B_x^2 + B_y^2}$<br /><br />Since $B_x = 0$, we have:<br /><br />$|\overrightarrow{B}| = |B_y|$<br /><br />Given that the magnitude of the resulting vector is 4, we have:<br /><br />$\sqrt{(4 - B_x)^2 + (-6 - B_y)^2} = 4$<br /><br />Substituting $B_x = 0$, we get:<br /><br />$\sqrt{4^2 + (-6 - B_y)^2} = 4$<br /><br />Squaring both sides, we get:<br /><br />$16 + (B_y + 6)^2 = 16$<br /><br />Simplifying, we get:<br /><br />$(B_y + 6)^2 = 0$<br /><br />Solving for $B_y$, we get:<br /><br />$B_y = -6$<br /><br />Therefore, the magnitude of $\overrightarrow{B}$ is:<br /><br />$|\overrightarrow{B}| = \sqrt{0^2 + (-6)^2} = 6$<br /><br />So, the correct answer is:<br /><br />A. 6