Sure! Let's find the sine of the angle between the vectors [tex]\(\vec{X} = 5\hat{i} - 3\hat{j} + 4\hat{k}\)[/tex] and [tex]\(\vec{Y} = \hat{j} - \hat{k}\)[/tex].
First, we need to find the dot product of [tex]\(\vec{X}\)[/tex] and [tex]\(\vec{Y}\)[/tex].
### Dot Product
The dot product of [tex]\(\vec{X}\)[/tex] and [tex]\(\vec{Y}\)[/tex] is given by:
[tex]\[
\vec{X} \cdot \vec{Y} = (5\hat{i} - 3\hat{j} + 4\hat{k}) \cdot (\hat{j} - \hat{k})
\][/tex]
The dot product is calculated as:
[tex]\[
\vec{X} \cdot \vec{Y} = 5 \cdot 0 + (-3) \cdot 1 + 4 \cdot (-1) = -3 - 4 = -7
\][/tex]
So, the dot product is [tex]\(\vec{X} \cdot \vec{Y} = -7\)[/tex].
### Magnitude of [tex]\(\vec{X}\)[/tex]
The magnitude of [tex]\(\vec{X}\)[/tex] is:
[tex]\[
|\vec{X}| = \sqrt{5^2 + (-3)^2 + 4^2} = \sqrt{25 + 9 + 16} = \sqrt{50} = 5\sqrt{2} \approx 7.071
\][/tex]
### Magnitude of [tex]\(\vec{Y}\)[/tex]
The magnitude of [tex]\(\vec{Y}\)[/tex] is:
[tex]\[
|\vec{Y}| = \sqrt{0^2 + 1^2 + (-1)^2} = \sqrt{0 + 1 + 1} = \sqrt{2} \approx 1.414
\][/tex]
### Cosine of the Angle
The cosine of the angle [tex]\(\theta\)[/tex] between the vectors [tex]\(\vec{X}\)[/tex] and [tex]\(\vec{Y}\)[/tex] is given by:
[tex]\[
\cos \theta = \frac{\vec{X} \cdot \vec{Y}}{|\vec{X}| \cdot |\vec{Y}|} = \frac{-7}{(5\sqrt{2}) \cdot \sqrt{2}} = \frac{-7}{10} = -0.7
\][/tex]
### Sine of the Angle
Finally, we use the identity [tex]\(\sin^2 \theta + \cos^2 \theta = 1\)[/tex] to find the sine of the angle.
[tex]\[
\sin^2 \theta = 1 - \cos^2 \theta = 1 - (-0.7)^2 = 1 - 0.49 = 0.51
\][/tex]
Taking the square root of both sides to get [tex]\(\sin \theta\)[/tex]:
[tex]\[
\sin \theta = \sqrt{0.51} \approx 0.714
\][/tex]
So, the sine of the angle between [tex]\(\vec{X}\)[/tex] and [tex]\(\vec{Y}\)[/tex] is approximately [tex]\(0.714\)[/tex].
