now that the sine of the angle $\theta$ between them is $\sin \theta = \sqrt{\frac{24}{28}}$.

Since $\sin^2 \theta + \cos^2 \theta = 1$, we can find $\cos \theta$ as follows:

$ \cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{\frac{1}{7}} $

The cosine of the angle $\theta$ is also given by:

$ \cos \theta = \frac{\hat{\mathbf{f}} \cdot \hat{\mathbf{g}}}{|\hat{\mathbf{f}}| \cdot |\hat{\mathbf{g}}|} $

Calculating the dot product:

$ \hat{\mathbf{f}} \cdot \hat{\mathbf{g}} = (1 \cdot 2) + (2 \cdot -3) + (-3 \cdot a) = 2 - 6 - 3a = -4 - 3a $

Calculating the magnitudes:

$ |\hat{\mathbf{f}}| = \sqrt{1^2 + 2^2 + (-3)^2} = \sqrt{14} $

$ |\hat{\mathbf{g}}| = \sqrt{2^2 + (-3)^2 + a^2} = \sqrt{13 + a^2} $

Thus, we have:

$ \frac{1}{\sqrt{7}} = \frac{-4 - 3a}{\sqrt{14} \cdot \sqrt{13 + a^2}} $

Simplifying, we find:

$ 2(13 + a^2) = (4 + 3a)^2 $

Expanding and rearranging gives:

$ 26 + 2a^2 = 16 + 9a^2 + 24a $

This simplifies to the quadratic equation:

$ 7a^2 + 24a - 10 = 0 $

Solving for $a$ using the quadratic formula:

$ a = \frac{-24 \pm \sqrt{24^2 - 4 \times 7 \times (-10)}}{2 \times 7} $

$ a = \frac{-24 \pm \sqrt{576 + 280}}{14} $

$ a = \frac{-24 \pm \sqrt{856}}{14} $

Finally, calculating $7a^2 + 24a$:

Substitute back into the quadratic result:

$ 7a^2 + 24a = 10 $

Vector Algebra

Continue this paper