Mathematics, 29.01.2020 02:08 babas97

If vector u has its initial point at (-7, 3) and its terminal point at (5, -6), u =

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Answer from: abbygriffin2009

First of all, let θθ be some angle in (0,π)(0,π). Then

θθ is acute ⟺⟺ θ<π2θ<π2 ⟺⟺ cosθ>0cos⁡θ>0.θθ is right ⟺⟺ θ=π2θ=π2 ⟺⟺ cosθ=0cos⁡θ=0.θθ is obtuse ⟺⟺ θ>π2θ>π2 ⟺⟺ cosθ<0cos⁡θ<0.

Now, to see if (say) angle AA of the triangle ABCABC is acute/right/obtuse, we need to check whether cos∠BACcos⁡∠BAC is positive/zero/negative. But what is cos∠BACcos⁡∠BAC? It is the angle made by the vectors AB−→−AB→ and AC−→−AC→. (When you are computing the angle at a particular vertex vv, you should make sure that both the vectors corresponding to the two adjacent sides have that vertex vv as the initial point.) We will first compute these two vectors:

AB−→−=(0,0,0)−(1,2,0)=(−1,−2,0)AB→=(0,0,0)−(1,2,0)=(−1,−2,0)AC−→−=(−2,1,0)−(1,2,0)=(−3,−1,0)AC→=(−2,1,0)−(1,2,0)=(−3,−1,0)Therefore, the angle between these vectors is given by:cos∠BAC=AB−→−⋅AC−→−|AB−→−||AC−→−|=…(1)(1)cos⁡∠BAC=AB→⋅AC→|AB→||AC→|=…Can you take it from here? From the sign of this value, you should be able to decide if angle AA is acute/right/obtuse.

Now, do the same procedure for the remaining two angles BB and CC as well. That should help you solve the problem.

A shortcut. Since you are not interested in the actual values of the angles, but you need only whether they are acute, obtuse or right, it is enough to compute only the sign of the numerator (the dot product between the vectors) in formula (1). The denominator is always positive.

Answer from: Quest

this may not be at all, but i use the same website you're using for school, and if i'm stuck doing one of these i try to look for it on the section.

Answer from: Quest

step-by-step explanation:

im not sure what the answer is

You are testing two scales for accuracy by weighing two different objects.use the drop-down menus to

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If vector u has its initial point at (-7, 3) and its terminal point at (5, -6), u =...