Answer :
To find the angle between the vectors [tex]\( u = \sqrt{5} t - 8 y \)[/tex] and [tex]\( v = \sqrt{5t} + y \)[/tex], we need to follow these steps:
1. Express the Vectors:
- For vector [tex]\( u \)[/tex], consider it as [tex]\( u = (\sqrt{5}t, -8y) \)[/tex].
- For vector [tex]\( v \)[/tex], consider it as [tex]\( v = (\sqrt{5}t, y) \)[/tex].
2. Calculate the Dot Product of [tex]\( u \)[/tex] and [tex]\( v \)[/tex]:
[tex]\[
u \cdot v = (\sqrt{5}t)(\sqrt{5}t) + (-8y)(y) = 5t^2 - 8y^2
\][/tex]
3. Calculate the Magnitude of Each Vector:
- Magnitude of [tex]\( u \)[/tex]:
[tex]\[
\|u\| = \sqrt{(\sqrt{5}t)^2 + (-8y)^2} = \sqrt{5t^2 + 64y^2}
\][/tex]
- Magnitude of [tex]\( v \)[/tex]:
[tex]\[
\|v\| = \sqrt{(\sqrt{5}t)^2 + (y)^2} = \sqrt{5t^2 + y^2}
\][/tex]
4. Find the Cosine of the Angle Between Them:
[tex]\[
\cos \theta = \frac{u \cdot v}{\|u\| \|v\|}
\][/tex]
Substituting the expressions we found:
[tex]\[
\cos \theta = \frac{5t^2 - 8y^2}{\sqrt{5t^2 + 64y^2} \cdot \sqrt{5t^2 + y^2}}
\][/tex]
5. Calculate the Angle:
To get the angle [tex]\(\theta\)[/tex], use the inverse cosine function:
[tex]\[
\theta = \cos^{-1}(\cos \theta)
\][/tex]
Convert the angle from radians to degrees and round to the nearest tenth of a degree.
From the solution process, the result we get is approximately:
[tex]\[
\boxed{98.5^\circ}
\][/tex]
So, the angle between the vectors [tex]\( u \)[/tex] and [tex]\( v \)[/tex] is closest to option B, which is [tex]\( 98.5^\circ \)[/tex].
1. Express the Vectors:
- For vector [tex]\( u \)[/tex], consider it as [tex]\( u = (\sqrt{5}t, -8y) \)[/tex].
- For vector [tex]\( v \)[/tex], consider it as [tex]\( v = (\sqrt{5}t, y) \)[/tex].
2. Calculate the Dot Product of [tex]\( u \)[/tex] and [tex]\( v \)[/tex]:
[tex]\[
u \cdot v = (\sqrt{5}t)(\sqrt{5}t) + (-8y)(y) = 5t^2 - 8y^2
\][/tex]
3. Calculate the Magnitude of Each Vector:
- Magnitude of [tex]\( u \)[/tex]:
[tex]\[
\|u\| = \sqrt{(\sqrt{5}t)^2 + (-8y)^2} = \sqrt{5t^2 + 64y^2}
\][/tex]
- Magnitude of [tex]\( v \)[/tex]:
[tex]\[
\|v\| = \sqrt{(\sqrt{5}t)^2 + (y)^2} = \sqrt{5t^2 + y^2}
\][/tex]
4. Find the Cosine of the Angle Between Them:
[tex]\[
\cos \theta = \frac{u \cdot v}{\|u\| \|v\|}
\][/tex]
Substituting the expressions we found:
[tex]\[
\cos \theta = \frac{5t^2 - 8y^2}{\sqrt{5t^2 + 64y^2} \cdot \sqrt{5t^2 + y^2}}
\][/tex]
5. Calculate the Angle:
To get the angle [tex]\(\theta\)[/tex], use the inverse cosine function:
[tex]\[
\theta = \cos^{-1}(\cos \theta)
\][/tex]
Convert the angle from radians to degrees and round to the nearest tenth of a degree.
From the solution process, the result we get is approximately:
[tex]\[
\boxed{98.5^\circ}
\][/tex]
So, the angle between the vectors [tex]\( u \)[/tex] and [tex]\( v \)[/tex] is closest to option B, which is [tex]\( 98.5^\circ \)[/tex].
