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Llana [10]
3 years ago
7

Vector question. Let v=<-2,1> u=<3,-5> , find:

Mathematics
1 answer:
lyudmila [28]3 years ago
5 0

Answer:

1. )\vec {v}.\vec{u}= -11

2.) The angle between v and u is 147.52°.

Step-by-step explanation:

Given:

\vec {v}= ( -2 , 1 )\\\vec {u}= ( 3 , -5 )

To Find:

1. \vec {v}.\vec{u}= ?

2.\theta = ?

Solution:

\vec {v}.\vec{u} is scalar product given as,

\vec {v}.\vec{u}=|\vec {v}||\vec {u}|\cos \theta

|\vec {v}|=|(-2, 1)|=\sqrt{(-2)^{2} +1^{2}}=\sqrt{5}\\|\vec {u}|=|(3, 5)|=\sqrt{(3)^{2} +(-5)^{2}}=\sqrt{34}

\vec {v}.\vec{u}=(-2i +j).(3i-5j)\\

Here only i.i = j.j =1 and i.j = j.i = 0

∴ \vec {v}.\vec{u}=(-2\times 3 +1\times -5)\\\vec {v}.\vec{u}=-6-5=-11 \\

Now, Substituting the above values we get

-11=\sqrt{5}\times \sqrt{34}\cos \theta\\ \cos \theta=\frac{-11}{\sqrt{170}} \\ \cos \theta =-0.84366\\\therefore \theta =cos^{-1}(-0.84366)\\\therefore \theta =147.52\°

As it is negative mean \theta is in Second Quadrant Because Cosine is negative in Second Quadrant.

1. )\vec {v}.\vec{u}= -11

2.) The angle between v and u is 147.52°.

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