WEBVTT
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Given the information in the diagram below, if ππ equals π times ππ, find π.
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In our diagram, we see that the bottom side of this triangle is parallel to the line segment that runs from point πΈ to point π·.
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Our problem statement tells us that this vector here, from point π΅ to point π·, is equal to some constant π times the vector from π· to π΄.
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Itβs π we want to solve for.
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And to do that, letβs rearrange this equation so π is the subject.
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Dividing both sides by vector ππ, this cancels on the right.
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And we see then that π equals ππ over ππ.
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Because these two vectors point in the same direction, we know that as we solve for this constant value π, itβs really the magnitudes or the lengths of these two vectors that weβre comparing.
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This means we can add vertical bars indicating magnitudes around the vectors of this expression.
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And from here we just read off these values from our diagram.
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Vector ππ has a magnitude of 4.8 centimeters, and ππ has a magnitude of 7.2 centimeters.
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Note that in this fraction, the units will cancel out.
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And 4.8 divided by 7.2 simplifies to two-thirds.
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This is the value by which we can multiply vector ππ so that itβs equal to vector ππ.