Key Concepts and Formulas

• Triangle Law of Vector Addition: For any three points A, B, and C, the vectors representing the sides of the triangle satisfy the relation $\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$. A more useful form for this problem is $\overrightarrow{CA} + \overrightarrow{AB} = \overrightarrow{CB}$.
• Area of a Triangle using Cross Product: The area of a triangle with adjacent sides represented by vectors $\vec{u}$ and $\vec{v}$ originating from the same vertex is given by $\frac{1}{2} |\vec{u} \times \vec{v}|$.
• Dot Product: For two vectors $\vec{A} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k}$ and $\vec{B} = B_x \hat{i} + B_y \hat{j} + B_z \hat{k}$, their dot product is $\vec{A} \cdot \vec{B} = A_x B_x + A_y B_y + A_z B_z$.

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Step-by-Step Solution

We are given the vectors: $\overrightarrow{\mathrm{AB}}=-2 \hat{i}+\hat{j}+3 \hat{k}$ $\overrightarrow{\mathrm{CB}}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$ $\overrightarrow{\mathrm{CA}}=4 \hat{i}+3 \hat{j}+\delta \hat{k}$ We are also given that $\delta > 0$ and the area of triangle ABC is $5 \sqrt{6}$. We need to find $\overrightarrow{CB} \cdot \overrightarrow{CA}$.

Step 1: Express $\overrightarrow{CB}$ in terms of $\overrightarrow{CA}$ and $\overrightarrow{AB}$

• Why this step? The components of $\overrightarrow{CB}$ are unknown ($\alpha, \beta, \gamma$). The Triangle Law of Vector Addition allows us to relate $\overrightarrow{CB}$ to the other given vectors, which will help us find these unknown components.
• Action: Apply the Triangle Law of Vector Addition: $\overrightarrow{CA} + \overrightarrow{AB} = \overrightarrow{CB}$.
• Working: Substituting the given vector expressions: $(4 \hat{i}+3 \hat{j}+\delta \hat{k}) + (-2 \hat{i}+\hat{j}+3 \hat{k}) = \alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$ Combining like terms on the left side: $(4-2) \hat{i} + (3+1) \hat{j} + (\delta+3) \hat{k} = \alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$ $2 \hat{i} + 4 \hat{j} + (\delta+3) \hat{k} = \alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$ Equating the coefficients of $\hat{i}$, $\hat{j}$, and $\hat{k}$, we get: $\alpha = 2$ $\beta = 4$ $\gamma = \delta+3$ Thus, $\overrightarrow{CB} = 2 \hat{i} + 4 \hat{j} + (\delta+3) \hat{k}$.

Step 2: Calculate the cross product of adjacent sides $\overrightarrow{CA}$ and $\overrightarrow{CB}$

• Why this step? The area of the triangle is given, and the formula for the area involves the magnitude of the cross product of two vectors representing adjacent sides. This step will provide an expression for the cross product that we can use with the given area.
• Action: Compute $\overrightarrow{CA} \times \overrightarrow{CB}$.
• Working: We have $\overrightarrow{CA} = 4 \hat{i}+3 \hat{j}+\delta \hat{k}$ and $\overrightarrow{CB} = 2 \hat{i}+4 \hat{j}+(\delta+3) \hat{k}$. The cross product is calculated using a determinant: $\overrightarrow{CA} \times \overrightarrow{CB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 3 & \delta \\ 2 & 4 & \delta+3 \end{vmatrix}$ $= \hat{i} [3(\delta+3) - 4\delta] - \hat{j} [4(\delta+3) - 2\delta] + \hat{k} [4 \cdot 4 - 2 \cdot 3]$ $= \hat{i} [3\delta+9 - 4\delta] - \hat{j} [4\delta+12 - 2\delta] + \hat{k} [16 - 6]$ $= \hat{i} (9-\delta) - \hat{j} (2\delta+12) + \hat{k} (10)$ So, $\overrightarrow{CA} \times \overrightarrow{CB} = (9-\delta) \hat{i} - (2\delta+12) \hat{j} + 10 \hat{k}$.

Step 3: Use the given area to find the value of $\delta$

• Why this step? We know the area of the triangle and have an expression for the cross product involving $\delta$. By equating the magnitude of half the cross product to the given area, we can form an equation to solve for $\delta$.
• Action: Set $\frac{1}{2} |\overrightarrow{CA} \times \overrightarrow{CB}| = 5\sqrt{6}$ and solve for $\delta$.
• Working: First, calculate the magnitude of the cross product: $|\overrightarrow{CA} \times \overrightarrow{CB}| = \sqrt{(9-\delta)^2 + (-(2\delta+12))^2 + (10)^2}$ $= \sqrt{(81 - 18\delta + \delta^2) + (4\delta^2 + 48\delta + 144) + 100}$ $= \sqrt{5\delta^2 + 30\delta + 325}$ Now, use the area formula: $\frac{1}{2} \sqrt{5\delta^2 + 30\delta + 325} = 5\sqrt{6}$ Multiply by 2: $\sqrt{5\delta^2 + 30\delta + 325} = 10\sqrt{6}$ Square both sides: $5\delta^2 + 30\delta + 325 = (10\sqrt{6})^2 = 100 \cdot 6 = 600$ Rearrange into a quadratic equation: $5\delta^2 + 30\delta + 325 - 600 = 0$ $5\delta^2 + 30\delta - 275 = 0$ Divide by 5: $\delta^2 + 6\delta - 55 = 0$ Factor the quadratic equation: $(\delta+11)(\delta-5) = 0$ The possible values for $\delta$ are $\delta = -11$ or $\delta = 5$. Since the problem states that $\delta > 0$, we choose $\delta = 5$. Now we can write the full vectors: $\overrightarrow{CA} = 4 \hat{i}+3 \hat{j}+5 \hat{k}$ $\overrightarrow{CB} = 2 \hat{i}+4 \hat{j}+(5+3) \hat{k} = 2 \hat{i}+4 \hat{j}+8 \hat{k}$

Step 4: Calculate the dot product $\overrightarrow{CB} \cdot \overrightarrow{CA}$

• Why this step? This is the final quantity we need to find, and we now have the complete vector expressions for $\overrightarrow{CB}$ and $\overrightarrow{CA}$.
• Action: Compute $\overrightarrow{CB} \cdot \overrightarrow{CA}$.
• Working: $\overrightarrow{CB} \cdot \overrightarrow{CA} = (2 \hat{i}+4 \hat{j}+8 \hat{k}) \cdot (4 \hat{i}+3 \hat{j}+5 \hat{k})$ $= (2)(4) + (4)(3) + (8)(5)$ $= 8 + 12 + 40$ $= 60$

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Common Mistakes & Tips

• Vector Direction: Ensure correct application of the Triangle Law of Vector Addition, paying attention to the order of vectors. For area calculation, vectors must originate from the same vertex.
• Cross Product Signs: Be careful with the sign of the $\hat{j}$ component when calculating the cross product determinant.
• Quadratic Solutions: Always verify that the solutions obtained for a variable satisfy any given constraints (e.g., $\delta > 0$).

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Summary

This problem requires combining vector addition, the cross product for area calculation, and the dot product. We first used the Triangle Law to express $\overrightarrow{CB}$ in terms of the other given vectors, allowing us to find its components in terms of $\delta$. Then, we used the formula for the area of a triangle involving the cross product to set up an equation and solve for $\delta$. Finally, with the complete vector components, we calculated the required dot product.

The final answer is $\boxed{60}$.