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# Help with vectors and that stuff.

+1
169
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+260

I got square root 113 for the first question.

How do i find the angle? do i to a tangent inverse?

Veteran  Mar 19, 2017
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#1
+79894
+1

tan (theta) =  (7 / -8)

arctan (7/-8) = theta  = -41.1859°

The terminal point of this vector lies in the second quadramt....so the angle is

180 - 41.1859  ≈ 138.81407° ≈  138. 8141°

And this is the angle ( theta ) that the vector makes with the positive x axis

CPhill  Mar 19, 2017
#2
+260
+1

That was one of my answers, the other one i got was like 84 degrees, i missed one class and it was about this stuff, so thanks. I will most likely ask a few more questions tonight if im stuck. I appriecate all the help around here.

Veteran  Mar 19, 2017
#3
+18777
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1. The magnitude of  $$\vec{u}$$.

$$\vec{u} = \binom{-8}{7}$$

$$\begin{array}{|rcll|} \hline ||\vec{v}|| &=& \sqrt{(-8)^2+7^2} \\ &=& \sqrt{64+49} \\ &=& \sqrt{113} \\ &=& 10.6301458127 \\ &\approx& 10.6301 \\ \hline \end{array}$$

2.The direction of $$\vec{u}$$, thaat is the angle $$\theta$$ it makes with the positive x-axis.

State your answer in degrees, rounded to at least four decimal places.

$$\vec{u} = \binom{-8}{7}$$

$$\begin{array}{|rcll|} \hline \tan(\theta) &=& \frac{|~\vec{e_x} \times \vec{u}~| } {\vec{e_x} \cdot \vec{v} } \\ &=& \frac{ \left|~\binom{1}{0} \times \binom{-8}{7}~\right| } {\binom{1}{0} \cdot \binom{-8}{7} } \\ &=& \frac{ (1)\cdot (7) - (0)\cdot (-8) } { (1)\cdot (-8) + (0)\cdot (7) } \\ &=& \frac{ 7+0 } { -8+0 } \\ &=& \frac{ 7} { -8 } \quad & | \quad II.\text{Quadrant} \\ \theta &=& \arctan(\frac{ 7 } { -8 }) \\ \theta &=& \arctan(-0.875) \\ \theta &=& -41.1859251657^{\circ} + 180^{\circ} \quad & | \quad II.\text{Quadrant} \\ \theta &=& 138.814074834^{\circ} \\ \theta &\approx& 138.8141^{\circ} \\ \hline \end{array}$$

heureka  Mar 20, 2017

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