Question:

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

Updated On: Nov 3, 2025
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Correct Answer: 3

Approach Solution - 1

To find the value of \( x \) such that the resultant of vectors \(\overrightarrow{OP}\), \(\overrightarrow{OQ}\), and \(\overrightarrow{OR}\) is \( A \sqrt{x} \), we must first analyze the vector configuration. Given that \(\angle POQ = 90^\circ\) and \(\angle QOR = 135^\circ\), use vector addition: 
\[ \overrightarrow{OP} = A \hat{i}, \quad \overrightarrow{OQ} = A \hat{j}, \quad \overrightarrow{OR} = -A \cos(45^\circ) \hat{i} - A \sin(45^\circ) \hat{j} \] \[ \overrightarrow{OR} = -A \frac{\sqrt{2}}{2} \hat{i} - A \frac{\sqrt{2}}{2} \hat{j} \] Now sum the vectors: \[ \overrightarrow{OP} + \overrightarrow{OQ} + \overrightarrow{OR} = \left(A - A \frac{\sqrt{2}}{2}\right) \hat{i} + \left(A - A \frac{\sqrt{2}}{2}\right) \hat{j} \] The resultant vector \(\overrightarrow{R}\) has the magnitude: \[ |\overrightarrow{R}| = \sqrt{\left(A - A \frac{\sqrt{2}}{2}\right)^2 + \left(A - A \frac{\sqrt{2}}{2}\right)^2} \] \[ = \sqrt{2 \left(A - A \frac{\sqrt{2}}{2}\right)^2} \] \[ = \sqrt{2 \left(A^2 \left(1-\frac{\sqrt{2}}{2}\right)^2\right)} \] \[ = A \sqrt{2 \left(1 - \frac{\sqrt{2}}{2}\right)^2} \] Simplifying, set the magnitude equal to \( A \sqrt{x} \), telling us: \[ A \sqrt{x}= A \sqrt{2 \left(1 - \frac{\sqrt{2}}{2}\right)^2} \] \[ x = 2 \left(1 - \frac{\sqrt{2}}{2}\right)^2 \] Calculate: \[ 1 - \frac{\sqrt{2}}{2} = \frac{2-\sqrt{2}}{2} \] \[ \left(\frac{2-\sqrt{2}}{2}\right)^2 = \frac{(2-\sqrt{2})^2}{4} = \frac{4 - 4\sqrt{2} + 2}{4} = \frac{6}{4} - \sqrt{2} + \frac{1}{2} \] Therefore, \[ x = 2 \times \left(\frac{3-2\sqrt{2}}{2}\right) = 3 \] Thus, the value of \( x \) is 3, which fits within the range (3, 3).

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Approach Solution -2

From the given diagram: Vectors $\overrightarrow{OP}$, $\overrightarrow{OQ}$, and $\overrightarrow{OR}$ form angles of $90^\circ$, $45^\circ$, and so on.
The resultant of the three vectors is:
\[\overrightarrow{R} = \overrightarrow{OP} + \overrightarrow{OQ} + \overrightarrow{OR}.\]
The magnitude is:
\[|\overrightarrow{R}| = \sqrt{\left(A + \frac{A}{\sqrt{2}}\right)^2 + \left(A + \frac{A}{\sqrt{2}}\right)^2}.\]
\[|\overrightarrow{R}| = \sqrt{\left(A + \frac{A}{\sqrt{2}}\right)^2 + \left(\frac{A}{\sqrt{2}}\right)^2}.\]
Simplify:
\[|\overrightarrow{R}| = A\sqrt{3}.\]
Thus, $x = 3$.
Final Answer: $x = 3$.

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