A vector vec Q which has a magnitude of 8 is | vedclass

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(B) Let $\vec P = P \hat{i}$ and $\vec Q = Q_x \hat{i} + Q_y \hat{j}$.Given,the resultant $\vec R = \vec P + \vec Q$ lies along the $y$-axis,so its $x

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$\frac {8}{\sqrt 5}$

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(B) Let $\vec P = P \hat{i}$ and $\vec Q = Q_x \hat{i} + Q_y \hat{j}$.Given,the resultant $\vec R = \vec P + \vec Q$ lies along the $y$-axis,so its $x$-component is zero.$R_x = P + Q_x = 0 \implies Q_x = -P$.The magnitude of $\vec Q$ is $8$,so $Q_x^2 + Q_y^2 = 8^2 = 64$.Substituting $Q_x = -P$,we get $P^2 + Q_y^2 = 64$.The resultant $\vec R$ lies along the $y$-axis,so $\vec R = R_y \hat{j} = (P + Q_y) \hat{j}$ (since $P+Q_x=0$,the $x$-component is zero).Given $|\vec R| = 2|\vec P| = 2P$,so $Q_y = 2P$.Substituting $Q_y = 2P$ into $P^2 + Q_y^2 = 64$:$P^2 + (2P)^2 = 64$$P^2 + 4P^2 = 64$$5P^2 = 64$$P^2 = \frac{64}{5}$$P = \sqrt{\frac{64}{5}} = \frac{8}{\sqrt{5}}$.

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

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