Question

Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}, \quad \vec{b} = -\hat{i} - 8\hat{j} + 2\hat{k}, \quad \text{and} \quad \vec{c} = 4\hat{i} + c_2\hat{j} + c_3\hat{k}$be three vectors such that $\vec{b} \times \vec{a} = \vec{c} \times \vec{a}.$If the angle between the vector $\vec{c}$ and the vector $3\hat{i} + 4\hat{j} + \hat{k}$ is $\theta$, then the greatest integer less than or equal to $\tan^2 \theta$ is:

Answer 1

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.

Answer 2

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.