Key Concepts and Formulas

• Vector Subtraction: For vectors $\vec{u} = u_x \hat{i} + u_y \hat{j} + u_z \hat{k}$ and $\vec{v} = v_x \hat{i} + v_y \hat{j} + v_z \hat{k}$, $\vec{u} - \vec{v} = (u_x - v_x) \hat{i} + (u_y - v_y) \hat{j} + (u_z - v_z) \hat{k}$.
• Area of a Triangle formed by two vectors: The area of a triangle with adjacent sides $\vec{u}$ and $\vec{v}$ is $\text{Area} = \frac{1}{2} |\vec{u} \times \vec{v}|$.
• Cross Product: The cross product of $\vec{u}$ and $\vec{v}$ is given by: $\vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ u_x & u_y & u_z \\ v_x & v_y & v_z \end{vmatrix}$
• Magnitude of a Vector: For a vector $\vec{v} = v_x \hat{i} + v_y \hat{j} + v_z \hat{k}$, its squared magnitude is $|\vec{v}|^2 = v_x^2 + v_y^2 + v_z^2$.

Step-by-Step Solution

Step 1: Determine the components of vector $\vec{c}$ We are given that $\vec{c} = \vec{a} - \vec{b}$. We perform vector subtraction component-wise. Given: $\overrightarrow{\mathrm{a}} = \alpha \hat{i} + 4 \hat{j} + 2 \hat{k}$ $\overrightarrow{\mathrm{b}} = 5 \hat{i} + 3 \hat{j} + 4 \hat{k}$ Subtracting $\vec{b}$ from $\vec{a}$: $\overrightarrow{\mathrm{c}} = (\alpha - 5) \hat{i} + (4 - 3) \hat{j} + (2 - 4) \hat{k}$ $\overrightarrow{\mathrm{c}} = (\alpha - 5) \hat{i} + 1 \hat{j} - 2 \hat{k}$ This gives us the components of $\vec{c}$ in terms of $\alpha$.

Step 2: Calculate the cross product $\vec{a} \times \vec{b}$ The area of the triangle formed by vectors $\vec{a}$ and $\vec{b}$ is given by $\frac{1}{2} |\vec{a} \times \vec{b}|$. We need to compute the cross product to use the given area. $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \alpha & 4 & 2 \\ 5 & 3 & 4 \end{vmatrix}$ Expanding the determinant: $\vec{a} \times \vec{b} = \hat{i} ((4)(4) - (2)(3)) - \hat{j} ((\alpha)(4) - (2)(5)) + \hat{k} ((\alpha)(3) - (4)(5))$ $\vec{a} \times \vec{b} = \hat{i} (16 - 6) - \hat{j} (4\alpha - 10) + \hat{k} (3\alpha - 20)$ $\vec{a} \times \vec{b} = 10 \hat{i} - (4\alpha - 10) \hat{j} + (3\alpha - 20) \hat{k}$

Step 3: Use the area of the triangle to form an equation for $\alpha$ We are given that the area of the triangle is $5 \sqrt{6}$. Using the formula for the area: $\text{Area} = \frac{1}{2} |\vec{a} \times \vec{b}|$ $5 \sqrt{6} = \frac{1}{2} |10 \hat{i} - (4\alpha - 10) \hat{j} + (3\alpha - 20) \hat{k}|$ Multiplying both sides by 2: $10 \sqrt{6} = |10 \hat{i} - (4\alpha - 10) \hat{j} + (3\alpha - 20) \hat{k}|$ Now, we square both sides to work with the squared magnitude: $(10 \sqrt{6})^2 = |10 \hat{i} - (4\alpha - 10) \hat{j} + (3\alpha - 20) \hat{k}|^2$ $100 \times 6 = (10)^2 + (-(4\alpha - 10))^2 + (3\alpha - 20)^2$ $600 = 100 + (4\alpha - 10)^2 + (3\alpha - 20)^2$ Subtract 100 from both sides: $500 = (4\alpha - 10)^2 + (3\alpha - 20)^2$ Expand the squared terms: $500 = (16\alpha^2 - 80\alpha + 100) + (9\alpha^2 - 120\alpha + 400)$ Combine like terms: $500 = (16\alpha^2 + 9\alpha^2) + (-80\alpha - 120\alpha) + (100 + 400)$ $500 = 25\alpha^2 - 200\alpha + 500$ Subtract 500 from both sides: $0 = 25\alpha^2 - 200\alpha$ Factor out $25\alpha$: $0 = 25\alpha (\alpha - 8)$ This gives two possible values for $\alpha$: $\alpha = 0$ or $\alpha = 8$.

Step 4: Select the correct value of $\alpha$ The problem states that $\alpha$ is a positive real number. Therefore, we discard $\alpha = 0$ and choose $\alpha = 8$.

Step 5: Calculate the squared magnitude of $\vec{c}$ Now that we have $\alpha = 8$, we can substitute this value back into the expression for $\vec{c}$ from Step 1. $\overrightarrow{\mathrm{c}} = (\alpha - 5) \hat{i} + 1 \hat{j} - 2 \hat{k}$ Substitute $\alpha = 8$: $\overrightarrow{\mathrm{c}} = (8 - 5) \hat{i} + 1 \hat{j} - 2 \hat{k}$ $\overrightarrow{\mathrm{c}} = 3 \hat{i} + 1 \hat{j} - 2 \hat{k}$ Finally, we calculate the squared magnitude of $\vec{c}$: $|\vec{c}|^2 = (3)^2 + (1)^2 + (-2)^2$ $|\vec{c}|^2 = 9 + 1 + 4$ $|\vec{c}|^2 = 14$

Common Mistakes & Tips

• Sign Errors in Cross Product: Be meticulous when calculating the determinant for the cross product, especially with the negative sign for the $\hat{j}$ component.
• Algebraic Expansion: Squaring binomials and combining terms in the quadratic equation can lead to errors. Double-check your algebraic manipulations.
• Positive Condition: Always remember to use all conditions given in the problem, such as $\alpha$ being positive, to select the correct solution.

Summary

We first determined the components of vector $\vec{c}$ using vector subtraction. Then, we computed the cross product of vectors $\vec{a}$ and $\vec{b}$. By equating half the magnitude of the cross product to the given area of the triangle, we formed a quadratic equation in terms of $\alpha$. Solving this equation and using the condition that $\alpha$ is positive, we found $\alpha = 8$. Finally, we substituted this value back into the expression for $\vec{c}$ and calculated its squared magnitude.

The final answer is $\boxed{\text{14}}$.