Question:
A vector \(\vec{P}\) directed along the \(x\)-axis is added to vector \(\vec{Q}\) which has a magnitude of \(10\,\text{m}\). The resultant vector is directed along the \(y\)-axis, with a magnitude that is \(2\) times that of \(\vec{P}\). The magnitude of \(\vec{P}\) is
A vector \(\vec{P}\) directed along the \(x\)-axis is added to vector \(\vec{Q}\) which has a magnitude of \(10\,\text{m}\). The resultant vector is directed along the \(y\)-axis, with a magnitude that is \(2\) times that of \(\vec{P}\). The magnitude of \(\vec{P}\) is
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When a resultant vector is along a coordinate axis, the component perpendicular to that axis must be zero. Use vector components and Pythagoras' theorem to relate the magnitudes.
Updated On: Jun 18, 2026
- \(\sqrt{10}\,\text{m}\)
- \(5\sqrt{2}\,\text{m}\)
- \(6\,\text{m}\)
- \(2\sqrt{5}\,\text{m}\)
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The Correct Option is D
Solution and Explanation
Step 1: Represent the vectors in component form.
Let the magnitude of \(\vec{P}\) be \(P\). Since \(\vec{P}\) is along the \(x\)-axis, \[ \vec{P}=(P,0). \] Let \[ \vec{Q}=(Q_x,Q_y). \] Given, \[ |\vec{Q}|=10. \] Hence, \[ Q_x^2+Q_y^2=100. \]
Step 2: Use the condition that the resultant is along the \(y\)-axis.
The resultant is \[ \vec{R}=\vec{P}+\vec{Q}. \] Since \(\vec{R}\) is directed along the \(y\)-axis, its \(x\)-component must be zero. Therefore, \[ P+Q_x=0. \] Thus, \[ Q_x=-P. \]
Step 3: Use the magnitude of the resultant.
The resultant is along the \(y\)-axis, so \[ |\vec{R}|=|Q_y|. \] Given that the magnitude of the resultant is twice that of \(\vec{P}\), \[ |Q_y|=2P. \] Hence, \[ Q_y^2=4P^2. \]
Step 4: Use the magnitude of \(\vec{Q}\).
Substituting \[ Q_x=-P \] and \[ Q_y^2=4P^2 \] into \[ Q_x^2+Q_y^2=100, \] we get \[ P^2+4P^2=100. \] \[ 5P^2=100. \] \[ P^2=20. \] \[ P=\sqrt{20}=2\sqrt{5}. \]
Step 5: Final conclusion.
Therefore, the magnitude of \(\vec{P}\) is \[ \boxed{2\sqrt{5}\,\text{m}} \]
Let the magnitude of \(\vec{P}\) be \(P\). Since \(\vec{P}\) is along the \(x\)-axis, \[ \vec{P}=(P,0). \] Let \[ \vec{Q}=(Q_x,Q_y). \] Given, \[ |\vec{Q}|=10. \] Hence, \[ Q_x^2+Q_y^2=100. \]
Step 2: Use the condition that the resultant is along the \(y\)-axis.
The resultant is \[ \vec{R}=\vec{P}+\vec{Q}. \] Since \(\vec{R}\) is directed along the \(y\)-axis, its \(x\)-component must be zero. Therefore, \[ P+Q_x=0. \] Thus, \[ Q_x=-P. \]
Step 3: Use the magnitude of the resultant.
The resultant is along the \(y\)-axis, so \[ |\vec{R}|=|Q_y|. \] Given that the magnitude of the resultant is twice that of \(\vec{P}\), \[ |Q_y|=2P. \] Hence, \[ Q_y^2=4P^2. \]
Step 4: Use the magnitude of \(\vec{Q}\).
Substituting \[ Q_x=-P \] and \[ Q_y^2=4P^2 \] into \[ Q_x^2+Q_y^2=100, \] we get \[ P^2+4P^2=100. \] \[ 5P^2=100. \] \[ P^2=20. \] \[ P=\sqrt{20}=2\sqrt{5}. \]
Step 5: Final conclusion.
Therefore, the magnitude of \(\vec{P}\) is \[ \boxed{2\sqrt{5}\,\text{m}} \]
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