Answer :
Sure! Let's find the angle between the vectors [tex]\( u = \sqrt{5} \, i - 8 \, j \)[/tex] and [tex]\( v = \sqrt{5} \, i + j \)[/tex].
1. Identify the components:
For vector [tex]\( u \)[/tex]:
- [tex]\( u_x = \sqrt{5} \)[/tex]
- [tex]\( u_y = -8 \)[/tex]
For vector [tex]\( v \)[/tex]:
- [tex]\( v_x = \sqrt{5} \)[/tex]
- [tex]\( v_y = 1 \)[/tex]
2. Calculate the dot product of [tex]\( u \)[/tex] and [tex]\( v \)[/tex]:
The dot product formula is:
[tex]\[
u \cdot v = u_x \times v_x + u_y \times v_y
\][/tex]
Substituting the values:
[tex]\[
u \cdot v = (\sqrt{5}) \times (\sqrt{5}) + (-8) \times 1 = 5 - 8 = -3
\][/tex]
3. Calculate the magnitudes of [tex]\( u \)[/tex] and [tex]\( v \)[/tex]:
The magnitude of [tex]\( u \)[/tex] is:
[tex]\[
\|u\| = \sqrt{u_x^2 + u_y^2} = \sqrt{(\sqrt{5})^2 + (-8)^2} = \sqrt{5 + 64} = \sqrt{69}
\][/tex]
The magnitude of [tex]\( v \)[/tex] is:
[tex]\[
\|v\| = \sqrt{v_x^2 + v_y^2} = \sqrt{(\sqrt{5})^2 + 1^2} = \sqrt{5 + 1} = \sqrt{6}
\][/tex]
4. Find the cosine of the angle [tex]\(\theta\)[/tex] between [tex]\( u \)[/tex] and [tex]\( v \)[/tex]:
The cosine formula is:
[tex]\[
\cos \theta = \frac{u \cdot v}{\|u\| \|v\|}
\][/tex]
Substitute the values we calculated:
[tex]\[
\cos \theta = \frac{-3}{\sqrt{69} \times \sqrt{6}}
\][/tex]
5. Compute the angle [tex]\( \theta \)[/tex] in degrees:
To find the angle [tex]\(\theta\)[/tex], we use the inverse cosine function:
[tex]\[
\theta = \cos^{-1}\left(\frac{-3}{\sqrt{69} \times \sqrt{6}}\right)
\][/tex]
After calculating, we find [tex]\(\theta\)[/tex] to be approximately 98.5 degrees.
Thus, the angle between the vectors [tex]\( u \)[/tex] and [tex]\( v \)[/tex] is approximately 98.5 degrees. The correct answer choice is b. 98.5.
1. Identify the components:
For vector [tex]\( u \)[/tex]:
- [tex]\( u_x = \sqrt{5} \)[/tex]
- [tex]\( u_y = -8 \)[/tex]
For vector [tex]\( v \)[/tex]:
- [tex]\( v_x = \sqrt{5} \)[/tex]
- [tex]\( v_y = 1 \)[/tex]
2. Calculate the dot product of [tex]\( u \)[/tex] and [tex]\( v \)[/tex]:
The dot product formula is:
[tex]\[
u \cdot v = u_x \times v_x + u_y \times v_y
\][/tex]
Substituting the values:
[tex]\[
u \cdot v = (\sqrt{5}) \times (\sqrt{5}) + (-8) \times 1 = 5 - 8 = -3
\][/tex]
3. Calculate the magnitudes of [tex]\( u \)[/tex] and [tex]\( v \)[/tex]:
The magnitude of [tex]\( u \)[/tex] is:
[tex]\[
\|u\| = \sqrt{u_x^2 + u_y^2} = \sqrt{(\sqrt{5})^2 + (-8)^2} = \sqrt{5 + 64} = \sqrt{69}
\][/tex]
The magnitude of [tex]\( v \)[/tex] is:
[tex]\[
\|v\| = \sqrt{v_x^2 + v_y^2} = \sqrt{(\sqrt{5})^2 + 1^2} = \sqrt{5 + 1} = \sqrt{6}
\][/tex]
4. Find the cosine of the angle [tex]\(\theta\)[/tex] between [tex]\( u \)[/tex] and [tex]\( v \)[/tex]:
The cosine formula is:
[tex]\[
\cos \theta = \frac{u \cdot v}{\|u\| \|v\|}
\][/tex]
Substitute the values we calculated:
[tex]\[
\cos \theta = \frac{-3}{\sqrt{69} \times \sqrt{6}}
\][/tex]
5. Compute the angle [tex]\( \theta \)[/tex] in degrees:
To find the angle [tex]\(\theta\)[/tex], we use the inverse cosine function:
[tex]\[
\theta = \cos^{-1}\left(\frac{-3}{\sqrt{69} \times \sqrt{6}}\right)
\][/tex]
After calculating, we find [tex]\(\theta\)[/tex] to be approximately 98.5 degrees.
Thus, the angle between the vectors [tex]\( u \)[/tex] and [tex]\( v \)[/tex] is approximately 98.5 degrees. The correct answer choice is b. 98.5.
