Key Concepts and Formulas

1. Vector Cross Product Property: If $\vec{A} \times \vec{B} = \vec{0}$ and $\vec{A}$ is a non-zero vector, then $\vec{B}$ must be parallel to $\vec{A}$, meaning $\vec{B} = \lambda \vec{A}$ for some scalar $\lambda$.
2. Distributive Property of Cross Product: $(\vec{X} - \vec{Y}) \times \vec{Z} = (\vec{X} \times \vec{Z}) - (\vec{Y} \times \vec{Z})$.
3. Angle Between Two Vectors: The cosine of the angle $\theta$ between two non-zero vectors $\vec{U}$ and $\vec{V}$ is given by $\cos \theta = \frac{\vec{U} \cdot \vec{V}}{|\vec{U}| |\vec{V}|}$.
4. Trigonometric Identity: The relationship between $\tan^2 \theta$ and $\cos^2 \theta$ is $\tan^2 \theta = \frac{1}{\cos^2 \theta} - 1$.

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Step-by-Step Solution

Step 1: Determine the unknown components of vector $\overrightarrow{c}$.

We are given the condition $\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}$. To utilize the properties of the cross product, we rearrange the equation: $\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}} - \overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}} = \overrightarrow{0}$ Using the distributive property of the cross product, we get: $(\overrightarrow{\mathrm{b}} - \overrightarrow{\mathrm{c}}) \times \overrightarrow{\mathrm{a}} = \overrightarrow{0}$ Since $\overrightarrow{\mathrm{a}} = \hat{i} + \hat{j} + \hat{k}$ is a non-zero vector, the vector $(\overrightarrow{\mathrm{b}} - \overrightarrow{\mathrm{c}})$ must be parallel to $\overrightarrow{\mathrm{a}}$. Therefore, we can write: $\overrightarrow{\mathrm{b}} - \overrightarrow{\mathrm{c}} = \lambda \overrightarrow{\mathrm{a}}$ for some scalar $\lambda$.

Now, we substitute the component forms of the vectors: $\overrightarrow{\mathrm{a}} = \hat{i} + \hat{j} + \hat{k}$ $\overrightarrow{\mathrm{b}} = -\hat{i} - 8 \hat{j} + 2 \hat{k}$ $\overrightarrow{\mathrm{c}} = 4 \hat{i} + \mathrm{c}_2 \hat{j} + \mathrm{c}_3 \hat{k}$

So, $(-\hat{i} - 8 \hat{j} + 2 \hat{k}) - (4 \hat{i} + \mathrm{c}_2 \hat{j} + \mathrm{c}_3 \hat{k}) = \lambda (\hat{i} + \hat{j} + \hat{k})$ Combining the terms on the left side: $(-1 - 4) \hat{i} + (-8 - \mathrm{c}_2) \hat{j} + (2 - \mathrm{c}_3) \hat{k} = \lambda \hat{i} + \lambda \hat{j} + \lambda \hat{k}$ $-5 \hat{i} + (-8 - \mathrm{c}_2) \hat{j} + (2 - \mathrm{c}_3) \hat{k} = \lambda \hat{i} + \lambda \hat{j} + \lambda \hat{k}$ Equating the coefficients of $\hat{i}$, $\hat{j}$, and $\hat{k}$ on both sides: For $\hat{i}$: $-5 = \lambda$ For $\hat{j}$: $-8 - \mathrm{c}_2 = \lambda$ For $\hat{k}$: $2 - \mathrm{c}_3 = \lambda$

From the $\hat{i}$ coefficient, we find $\lambda = -5$. Substituting $\lambda = -5$ into the other equations: $-8 - \mathrm{c}_2 = -5 \implies \mathrm{c}_2 = -8 + 5 \implies \mathrm{c}_2 = -3$ $2 - \mathrm{c}_3 = -5 \implies \mathrm{c}_3 = 2 + 5 \implies \mathrm{c}_3 = 7$

Thus, the vector $\overrightarrow{\mathrm{c}}$ is $4 \hat{i} - 3 \hat{j} + 7 \hat{k}$.

Step 2: Calculate the angle $\theta$ between $\overrightarrow{c}$ and the vector $3 \hat{i}+4 \hat{j}+\hat{k}$.

Let $\overrightarrow{\mathrm{d}} = 3 \hat{i} + 4 \hat{j} + \hat{k}$. We need to find the angle $\theta$ between $\overrightarrow{\mathrm{c}} = 4 \hat{i} - 3 \hat{j} + 7 \hat{k}$ and $\overrightarrow{\mathrm{d}}$.

First, calculate the dot product $\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{d}}$: $\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{d}} = (4)(3) + (-3)(4) + (7)(1) = 12 - 12 + 7 = 7$ Next, calculate the magnitudes of $\overrightarrow{\mathrm{c}}$ and $\overrightarrow{\mathrm{d}}$: $|\overrightarrow{\mathrm{c}}| = \sqrt{4^2 + (-3)^2 + 7^2} = \sqrt{16 + 9 + 49} = \sqrt{74}$ $|\overrightarrow{\mathrm{d}}| = \sqrt{3^2 + 4^2 + 1^2} = \sqrt{9 + 16 + 1} = \sqrt{26}$ Now, use the formula for the cosine of the angle between two vectors: $\cos \theta = \frac{\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{d}}}{|\overrightarrow{\mathrm{c}}| |\overrightarrow{\mathrm{d}}|} = \frac{7}{\sqrt{74} \sqrt{26}} = \frac{7}{\sqrt{1924}}$

Step 3: Calculate $\tan^2 \theta$ and find the greatest integer less than or equal to it.

We have $\cos \theta = \frac{7}{\sqrt{1924}}$. First, calculate $\cos^2 \theta$: $\cos^2 \theta = \left( \frac{7}{\sqrt{1924}} \right)^2 = \frac{49}{1924}$ Now, use the trigonometric identity $\tan^2 \theta = \frac{1}{\cos^2 \theta} - 1$: $\tan^2 \theta = \frac{1}{\frac{49}{1924}} - 1 = \frac{1924}{49} - 1$ $\tan^2 \theta = \frac{1924 - 49}{49} = \frac{1875}{49}$ To find the greatest integer less than or equal to $\tan^2 \theta$, we perform the division: $\frac{1875}{49} \approx 38.2653$ The greatest integer less than or equal to $38.2653$ is $38$.

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Common Mistakes & Tips

• Ensure correct handling of signs during vector subtraction and dot product calculations.
• Double-check the calculation of magnitudes of vectors.
• Remember the identity $\tan^2 \theta = \sec^2 \theta - 1$ and its relation to $\cos^2 \theta$.
• The greatest integer function $\lfloor x \rfloor$ rounds down to the nearest integer.

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Summary

The problem required us to first determine the unknown components of vector $\overrightarrow{\mathrm{c}}$ using the property that if $\vec{X} \times \vec{Y} = \vec{0}$, then $\vec{X}$ and $\vec{Y}$ are parallel. This led to $\overrightarrow{\mathrm{b}} - \overrightarrow{\mathrm{c}} = \lambda \overrightarrow{\mathrm{a}}$, allowing us to solve for $\mathrm{c}_2$ and $\mathrm{c}_3$. Subsequently, we calculated the cosine of the angle $\theta$ between $\overrightarrow{\mathrm{c}}$ and the given vector $\overrightarrow{\mathrm{d}}$ using the dot product formula. Finally, we used a trigonometric identity to find $\tan^2 \theta$ and then determined the greatest integer less than or equal to its value.

The final answer is $\boxed{38}$.