Question

PLease help! timedd

Find the angle between u = sqr5i-8J and v = sqr5i+j. Round to the nearest tenth of a degree. (the square root is only on the 5i in both)
A.65.9
B.98.5
C.90.4
D33.3



Use the dot product to find [v] when v =(-2,-1)


A.-1

b.-3

c;sqr5

dsqr3

Answer 1

98.5 degrees

Explanation:

To find the angle measurement between two vectors use: cos(theta)=(u dot v)/(|u|*|v|)

The vectors are u=sqrt(5)i-8j and v=sqrt(5)i+1j .

So find the dot product of u and v. u dot v=sqrt5*sqrt(5)+(-8)*(1)=5-8=-3.

Second step: Find the magnitude of both vectors. So to find the magnitude of a vector, let's call it t when t=ai+bj, you just do |t|=sqrt(a^2+b^2).

|u|=sqrt(5+64)=sqrt(69) and |v|=sqrt(5+1)=sqrt(6)

Now multiply your magnitudes of your vectors: sqrt(69)*sqrt(6)=sqrt(414).

So now we have:

cos(theta)=-3/sqrt(414)

Now arccos( ) or cos^(-1) to find the angle theta.

theta=cos^(-1)[-3/sqrt(414)] which is approximately 98.5 degrees.

Answer 2

Ans 1) The correct option is B) 98.5

Ans 2) The correct option is C)
`√(5)`

Explanation:

The angle measurement between two vectors by:

`cos(\theta)=((u.v))/((|u|* |v|))`

Magnitude of vector t=ai+bj, calculated by:

`|t|=\sqrt{a^(2)+b^(2)}`

The given vectors are
`u=√(5)i-8j` and
`v=√(5)i+1j`

First find the dot product of u and v;

`u.v=(√(5)i-8j)(√(5)i+1j) = 5-8= - 3`

Now, Find the magnitude of both vectors u and v.

`|u|=\sqrt{(√(5))^(2)+8^(2)}`

`|u|=√(5+64)`

`|u|=√(69)`

and

`|v|=\sqrt{(√(5))^(2)+1^(2)}`

`|v|=√(5+1)`

`|v|=√(6)`

Now, put the all values in
`cos(\theta)=((u.v))/((|u|* |v|))`

`cos(\theta)=(-3)/(√(69) * √(6))`

`=(-3)/(414)`

take arc cos both the sides,

`\theta=cos^(-1) (-3)/(\414)`

`\theta=98.5 \degree` (approx)

Therefore, the correct option is B) 98.5

Ans 2) calculate IvI by

If t= ai +bj then magnitute is
`\sqrt{a^(2)+b^(2)}`

Given : v = (-2,-1)

it means v = -2i -1 j

`|v| =\sqrt{(-2)^(2)+(-1)^(2)}`

`|v| =√(4+1)`

`|v| =√(5)`

Therefore, the correct option is C)
`√(5)`