How do you find the component form and magnitude of the vector v given initial point (-1,5) and terminal point (15, 12)?

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Answer 1 · 2017-03-24T23:00:02

Component form is v $= < 16, 7 >$ $= <16, 7>$
Magnitude is ||v|| = $\sqrt{305} \approx 17.46$ $sqrt305~~ 17.46$

To find the component form, you only need to know how to substitute figures for letters. What do I mean by this?

If initial side is $(x_{1}, y_{1})$ $(x_1, y_1)$

Then $x_{1} = - 1$ $x_1 =-1$ and $y_{1} = 5$ $y_1 = 5$

If terminal side is $(x_{2}, y_{2})$ $(x_2, y_2)$

Then $x_{2} = 15$ $x_2 = 15$ and $y_{2} = 12$ $y_2 = 12$

Thus, component form of v is $< (x_{2} - x_{1}), (y_{2} - y_{1}) >$ $< (x_2 - x_1), (y_2 - y_1) >$ ...simply $< x, y >$ $<x, y>$

In this case, v $= < [15 - (- 1)], (12 - 5)$ $= < [15 - (-1)], (12 - 5)$

Which gives us v $= < 16, 7 >$ $= <16, 7>$

To find the magnitude of a vector, the concept of pythagorean theorem needs to be understood. Why?

If v $= < x, y >$ $= <x, y>$ Using pythagorean theorem, ||v|| $= \sqrt{x^{2} + y^{2}}$ $= sqrt(x^2+ y^2)$

We already know the values for v, so;
||v|| = $\sqrt{16^{2} + 7^{2}}$ $sqrt(16^2+7^2$

||v|| = $\sqrt{256 + 49}$ $sqrt(256+49)$

||v|| = $\sqrt{305}$ $sqrt305$

||v|| $\approx 17.46$ $~~ 17.46$