Calculate $\vec{b} \times \vec{a}$:
\(\vec{b} \times \vec{a}\) = \(\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1 & -8 & 21 \ 1 & 1 & 1 \end{vmatrix} = -10\hat{i} + 3\hat{j} + 7\hat{k}\)
Since \(\vec{b} \times \vec{a} = \vec{c} \times \vec{a},\) we have:
$-10\hat{i} + 3\hat{j} + 7\hat{k} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ c_1 & c_2 & c_3 \ 1 & 1 & 1 \end{vmatrix}$
Expanding the determinant, we get:
$-10\hat{i} + 3\hat{j} + 7\hat{k} = (c_2 - c_3)\hat{i} - (c_1 - c_3)\hat{j} + (c_1 - c_2)\hat{k}$
Comparing the coefficients, we get:
$c_2 - c_3 = -10$
$-c_1 + c_3 = -3$
$c_1 - c_2 = 7$
Solving these equations, we find:
$c_2 = -3$
$c_3 = 7$
$c_1 = 4$
So, $\vec{c} = 4\hat{i} - 3\hat{j} + 7\hat{k}$.
Let $\theta$ be the angle between the two vectors. We can use the dot product formula:
$(3\hat{i} + 4\hat{j} + \hat{k}) \cdot \vec{c} = |\vec{c}||3\hat{i} + 4\hat{j} + \hat{k}| \cos \theta$
Calculating the dot product and magnitudes:
$(4,-3,7) \cdot (3,4,1) = \sqrt{74}\sqrt{26} \cos \theta$
Simplifying:
$12 - 12 + 7 = \sqrt{74}\sqrt{26} \cos \theta$
$7 = \sqrt{74}\sqrt{26} \cos \theta$
Solving for $\cos \theta$:
$\cos \theta = \frac{7}{\sqrt{74}\sqrt{26}}$
Using the identity $\sin^2 \theta + \cos^2 \theta = 1$, we can find $\sin \theta$:
$\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - \frac{49}{1924}} = \frac{\sqrt{1875}}{1924}$
Now, we can calculate $\tan^2 \theta$:
$\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} = \frac{1875}{49}$
The greatest integer less than or equal to $\frac{1875}{49}$ is 38.
Therefore, the correct answer is 38.
A body of mass 1000 kg is moving horizontally with a velocity of 6 m/s. If 200 kg extra mass is added, the final velocity (in m/s) is: