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

Correct Answer: 38

To solve the given problem, we start by analyzing the condition \(\vec{b} \times \vec{a} = \vec{c} \times \vec{a}\). Given the vectors: \(\vec{a} = \hat{i} + \hat{j} + \hat{k}\), \(\vec{b} = -\hat{i} - 8\hat{j} + 2\hat{k}\), and \(\vec{c} = 4\hat{i} + c_2\hat{j} + c_3\hat{k}\), the condition implies that \((\vec{b} - \vec{c}) \times \vec{a} = \vec{0}\). This means \(\vec{b} - \vec{c}\) is parallel to \(\vec{a}\).
Calculating \(\vec{b} - \vec{c}\):
\(\vec{b} - \vec{c} = (-1 - 4)\hat{i} + (-8 - c_2)\hat{j} + (2 - c_3)\hat{k} = -5\hat{i} + (-8 - c_2)\hat{j} + (2 - c_3)\hat{k}\).
Since \(\vec{b} - \vec{c}\) is parallel to \(\vec{a}\), it must be a scalar multiple: \(-5\hat{i} + (-8 - c_2)\hat{j} + (2 - c_3)\hat{k} = \lambda(\hat{i} + \hat{j} + \hat{k})\).
Equating components, we get:
- \(-5 = \lambda\)
- \(-8 - c_2 = \lambda\)
- \(2 - c_3 = \lambda\)
Solving these equations:
- From \(-5 = \lambda\), we have \(\lambda = -5\).
- Plugging \(\lambda = -5\) into \(-8 - c_2 = \lambda\):
\(-8 - c_2 = -5 \Rightarrow c_2 = -3\).
- Plugging \(\lambda = -5\) into \(2 - c_3 = \lambda\):
\(2 - c_3 = -5 \Rightarrow c_3 = 7\).
Thus, \(\vec{c} = 4\hat{i} - 3\hat{j} + 7\hat{k}\).
Next, consider the angle \(\theta\) between \(\vec{c}\) and \(3\hat{i} + 4\hat{j} + \hat{k}\).
The cosine of the angle is given by the formula:
\[\cos \theta = \frac{\vec{c} \cdot (3\hat{i} + 4\hat{j} + \hat{k})}{|\vec{c}||3\hat{i} + 4\hat{j} + \hat{k}|}\]
Calculating the dot product \(\vec{c} \cdot (3\hat{i} + 4\hat{j} + \hat{k})\):
\(= (4 \cdot 3) + (-3 \cdot 4) + (7 \cdot 1) = 12 - 12 + 7 = 7\).
Finding the magnitudes:
- \(|\vec{c}| = \sqrt{4^2 + (-3)^2 + 7^2} = \sqrt{16 + 9 + 49} = \sqrt{74}\).
- \(|3\hat{i} + 4\hat{j} + \hat{k}| = \sqrt{3^2 + 4^2 + 1^2} = \sqrt{9 + 16 + 1} = \sqrt{26}\).
Thus, \(\cos \theta = \frac{7}{\sqrt{74} \cdot \sqrt{26}} = \frac{7}{\sqrt{1924}}\).
Using \(\cos^2 \theta + \sin^2 \theta = 1\), find \(\sin^2 \theta\):
\(\sin^2 \theta = 1 - \cos^2 \theta = 1 - \left(\frac{7^2}{1924}\right) = 1 - \frac{49}{1924} = \frac{1924 - 49}{1924} = \frac{1875}{1924}\).
Then calculate \(\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta}\):
\(\tan^2 \theta = \frac{\frac{1875}{1924}}{\frac{49}{1924}} = \frac{1875}{49} \approx 38.2653\).
The greatest integer less than or equal to \(\tan^2 \theta\) is \(38\), which matches the expected range (38,38).

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.