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(D) Let the vector $\overrightarrow{OP}$ be along the $x$-axis,$\overrightarrow{OP} = A \hat{i}$.Vector $\overrightarrow{OQ}$ is at $90^\circ$ to $\ov

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) Let the vector $\overrightarrow{OP}$ be along the $x$-axis,$\overrightarrow{OP} = A \hat{i}$.Vector $\overrightarrow{OQ}$ is at $90^\circ$ to $\overrightarrow{OP}$,so $\overrightarrow{OQ} = A \hat{j}$.Vector $\overrightarrow{OR}$ is at $45^\circ$ below the $x$-axis,so $\overrightarrow{OR} = A \cos(45^\circ) \hat{i} - A \sin(45^\circ) \hat{j} = \frac{A}{\sqrt{2}} \hat{i} - \frac{A}{\sqrt{2}} \hat{j}$.The resultant vector $\vec{R} = \overrightarrow{OP} + \overrightarrow{OQ} + \overrightarrow{OR} = (A + \frac{A}{\sqrt{2}}) \hat{i} + (A - \frac{A}{\sqrt{2}}) \hat{j}$.The magnitude of the resultant is $|\vec{R}| = \sqrt{(A + \frac{A}{\sqrt{2}})^2 + (A - \frac{A}{\sqrt{2}})^2}$.$|\vec{R}| = \sqrt{A^2(1 + \frac{1}{\sqrt{2}})^2 + A^2(1 - \frac{1}{\sqrt{2}})^2} = A \sqrt{(1 + \frac{1}{2} + \sqrt{2}) + (1 + \frac{1}{2} - \sqrt{2})} = A \sqrt{1 + 0.5 + 1 + 0.5} = A \sqrt{3}$.Comparing $A \sqrt{3}$ with $A \sqrt{x}$,we get $x = 3$.

Three vectors $\overrightarrow{OP}, \overrightarrow{OQ}$ and $\overrightarrow{OR}$ each of magnitude $A$ are acting as shown in the figure. The resultant of the three vectors is $A \sqrt{x}$. The value of $x$ is:

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