<h2 class="title">Using Pythagoras' Theorem to Find Magnitude</h2>
<div class="wp-caption alignnone" id="attachment_103643" style="width: 536px"><a href="https://data.ulearngo.com/assets/f4364743-4435-4d8e-9c51-48b06cfb53ca"><img alt="resultant-vector" class="wp-image-103643 size-full" height="359" src="https://data.ulearngo.com/assets/f4364743-4435-4d8e-9c51-48b06cfb53ca" width="536"/></a><p class="wp-caption-text">Adding co-linear vectors to get a resultant vector.</p></div>
<p>If we wanted to know the resultant of the three blue vectors and the three red vectors in the figure above we can use the resultant vectors in the \(x\)- and \(y\)-directions to determine this.</p>
<div class="wp-caption alignnone" style="width: 367px"><img alt="c65a486766bd3b90f2c3bd72016967ff.png" height="400" src="https://data.ulearngo.com/assets/4abcea0c-07f5-41e7-b73f-b82fca89561b" width="367"/><p class="wp-caption-text">Finding the resultant.</p></div>
<p>The black arrow represents the resultant of the vectors \(\vec{R}_x\) and \(\vec{R}_y\). We can find the magnitude of this vector using the theorem of Pythagoras because the three vectors form a right angle triangle. If we had drawn the vectors to scale we would be able to measure the magnitude of the resultant as well.</p>
<p>What we've actually sketched out already is our approach to finding the resultant of many vectors using components so remember this example when we get there a little later.</p>
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<h2>Example: Finding the Magnitude of the Resultant</h2>
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<h3>Question</h3>
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<p>The force vectors in the <a href="#attachment_103643">figure above</a> have the following magnitudes: \(\text{1}\) \(\text{N}\), \(\text{1}\) \(\text{N}\), \(\text{2}\) \(\text{N}\) for the blue ones and \(\text{2}\) \(\text{N}\), \(\text{2}\) \(\text{N}\) and \(\text{1.5}\) \(\text{N}\) for the red ones. Determine the magnitude of the resultant.</p>
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<h3>Step 1: Determine the resultant of the vectors parallel to the \(y\)-axis</h3>
<p>The resultant of the vectors parallel to the \(y\)-axis is found by adding the magnitudes (lengths) of three vectors because they all point in the same direction. The answer is \(\vec{R}_y\)=\(\text{1}\) \(\text{N}\) + \(\text{1}\) \(\text{N}\) + \(\text{2}\) \(\text{N}\) = \(\text{4}\) \(\text{N}\) in the positive \(y\)-direction.</p>
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<h3>Step 2: Determine the resultant of the vectors parallel to the \(x\)-axis</h3>
<p>The resultant of the vectors parallel to the \(x\)-axis is found by adding the magnitudes (lengths) of three vectors because they all point in the same direction. The answer is \(\vec{R}_x\)=\(\text{2}\) \(\text{N}\) + \(\text{2}\) \(\text{N}\) + \(\text{1.5}\) \(\text{N}\) = \(\text{5.5}\) \(\text{N}\) in the positive \(x\)-direction.</p>
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<h3>Step 3: Determine the magnitude of the resultant</h3>
<p>We have a right angled triangle. We also know the length of two of the sides. Using Pythagoras we can find the length of the third side. From what we know about resultant vectors this length will be the magnitude of the resultant vector.</p>
<p>The resultant is: \begin{align*} R_x^{2} + R_y^{2} &amp;= R^{2}\ \text{(Pythagoras' theorem)}\\ (\text{5,5})^{2} + (4)^{2} &amp;= R^{2}\\ R &amp;= \text{6,8} \end{align*}</p>
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<h3>Step 4: Quote the final answer</h3>
<p>Magnitude of the resultant: \(\text{6.8}\) \(\text{N}\)</p>
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<p><strong>Note:</strong> we did not determine the resultant vector in the worked example above because we only determined the magnitude. A vector needs a <strong>magnitude</strong> and a <strong>direction</strong>. We did not determine the direction of the resultant vector.</p>

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Finding Magnitude With Pythagoras Theorem

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Finding Magnitude With Pythagoras Theorem

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