Answer:
98.5 degrees
Step-by-step 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:
Ans 1) The correct option is B) 98.5
Ans 2) The correct option is C) [tex]\sqrt{5}[/tex]
Step-by-step explanation:
The angle measurement between two vectors by:
[tex]cos(\theta)=\frac{(u.v)}{(|u|\times |v|)}[/tex]
Magnitude of vector t=ai+bj, calculated by:
[tex]|t|=\sqrt{a^{2}+b^{2}}[/tex]
The given vectors are [tex]u=\sqrt{5}i-8j[/tex] and [tex]v=\sqrt{5}i+1j[/tex]
First find the dot product of u and v;
[tex]u.v=(\sqrt{5}i-8j)(\sqrt{5}i+1j) = 5-8= - 3[/tex]
Now, Find the magnitude of both vectors u and v.
[tex]|u|=\sqrt{(\sqrt{5})^{2}+8^{2}}[/tex]
[tex]|u|=\sqrt{5+64}[/tex]
[tex]|u|=\sqrt{69}[/tex]
and
[tex]|v|=\sqrt{(\sqrt{5})^{2}+1^{2}}[/tex]
[tex]|v|=\sqrt{5+1}[/tex]
[tex]|v|=\sqrt{6}[/tex]
Now, put the all values in [tex]cos(\theta)=\frac{(u.v)}{(|u|\times |v|)}[/tex]
[tex]cos(\theta)=\frac{-3}{\sqrt{69} \times \sqrt{6}}[/tex]
[tex]=\frac{-3}{414}[/tex]
take arc cos both the sides,
[tex]\theta=cos^{-1} \frac{-3}{\414}[/tex]
[tex]\theta=98.5 \degree[/tex] (approx)
Therefore, the correct option is B) 98.5
Ans 2) calculate IvI by
If t= ai +bj then magnitute is [tex]\sqrt{a^{2}+b^{2}}[/tex]
Given : v = (-2,-1)
it means v = -2i -1 j
[tex]|v| =\sqrt{(-2)^{2}+(-1)^{2}}[/tex]
[tex]|v| =\sqrt{4+1}[/tex]
[tex]|v| =\sqrt{5}[/tex]
Therefore, the correct option is C) [tex]\sqrt{5}[/tex]
