Let for a triangle ABC AB = -2i + j + 3k, CB = αi + βj + γk, CA = 4i + 3j + δk. If δ 0 and the are

Q: Wait, why is the Triangle Law $\vec{CA} + \vec{AB} = \vec{CB}$ and not something else?

Think of vectors as displacements. If you start at point $C$, walk to $A$ (vector $\vec{CA}$), and then walk from $A$ to $B$ (vector $\vec{AB}$), your final position relative to your starting point is $B$. Therefore, your total displacement is the vector from $C$ to $B$, which is $\vec{CB}$. It is a simple path-following logic!

Q: Why do we use the cross product for the area of a triangle?

The cross product of two vectors $\vec{a}$ and $\vec{b}$ produces a vector whose magnitude is equal to the area of the parallelogram formed by $\vec{a}$ and $\vec{b}$. Since a triangle is exactly half of that parallelogram, the area is $\frac{1}{2} |\vec{a} \times \vec{b}|$. It is a powerful geometric property that turns a 3D vector problem into a manageable algebraic one.

Q: Why did we reject $\delta = -11$?

In mathematics, we often find multiple roots for a quadratic equation, but physics and geometry impose constraints. The problem explicitly stated $\delta > 0$. This is a physical constraint on the geometry of the triangle. Always check your final values against the initial conditions provided in the question!

Question

Let for a triangle $ABC$ $\vec{AB} = -2\hat{i} + \hat{j} + 3\hat{k}$, $\vec{CB} = \alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}$, $\vec{CA} = 4\hat{i} + 3\hat{j} + \delta\hat{k}$. If $\delta > 0$ and the area of the triangle $ABC$ is $5\sqrt{6}$ then $\vec{CB} \cdot \vec{CA}$ is equal to

Options:

(A) 60

(B) 54

(C) 108

(D) 120

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional coordinate system. You have three points, $A$, $B$, and $C$, floating in space, connected by vectors.

We are given the following vectors:
$\vec{AB} = -2\hat{i} + \hat{j} + 3\hat{k}$
$\vec{CB} = \alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}$
$\vec{CA} = 4\hat{i} + 3\hat{j} + \delta\hat{k}$

Our mission is to find the dot product $\vec{CB} \cdot \vec{CA}$, but to get there, we must first unlock the secret of the unknown parameter $\delta$.

### The Triangle Law

The first step in any vector journey is understanding the connectivity. If you start at point $C$, travel to $A$ along $\vec{CA}$, and then travel from $A$ to $B$ along $\vec{AB}$, your net displacement is the vector $\vec{CB}$.

This is the Triangle Law of Vector Addition:
$$\vec{CA} + \vec{AB} = \vec{CB}$$

By substituting our known components, we get:
$$(4\hat{i} + 3\hat{j} + \delta\hat{k}) + (-2\hat{i} + \hat{j} + 3\hat{k}) = \alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}$$

Comparing the components, we find $\alpha = 2$, $\beta = 4$, and $\gamma = \delta + 3$. We have successfully expressed $\vec{CB}$ in terms of $\delta$.

### The Area Constraint

Now, we turn to the area. We know the area of $\Delta ABC$ is $5\sqrt{6}$. The area of a triangle formed by vectors $\vec{AB}$ and $\vec{CA}$ is given by the formula:
$$	ext{Area} = \frac{1}{2} |\vec{AB} 	imes \vec{CA}|$$

Multiplying by $2$, we get $|\vec{AB} 	imes \vec{CA}| = 10\sqrt{6}$. To find this cross product, we use the determinant method:
$$\vec{AB} 	imes \vec{CA} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -2 & 1 & 3 \ 4 & 3 & \delta \end{vmatrix}$$

Expanding this determinant, we get:
$$\vec{AB} 	imes \vec{CA} = \hat{i}(\delta - 9) - \hat{j}(-2\delta - 12) + \hat{k}(-6 - 4) = (\delta - 9)\hat{i} + (2\delta + 12)\hat{j} - 10\hat{k}$$

### The Algebraic Grind

We now have the cross product vector. Its magnitude squared must equal $(10\sqrt{6})^2 = 600$.

So:
$$(\delta - 9)^2 + (2\delta + 12)^2 + (-10)^2 = 600$$

Expanding these terms:
$$(\delta^2 - 18\delta + 81) + (4\delta^2 + 48\delta + 144) + 100 = 600$$

Combining like terms leads us to:
$$5\delta^2 + 30\delta + 325 = 600 \implies 5\delta^2 + 30\delta - 275 = 0$$

Dividing by $5$, we arrive at the elegant quadratic:
$$\delta^2 + 6\delta - 55 = 0$$

Factoring this, we get $(\delta + 11)(\delta - 5) = 0$. Since the problem dictates $\delta > 0$, we must reject $-11$ and accept $\delta = 5$.

### Final Calculation

With $\delta = 5$, our vectors are fully defined:
$\vec{CA} = 4\hat{i} + 3\hat{j} + 5\hat{k}$
$\vec{CB} = 2\hat{i} + 4\hat{j} + 8\hat{k}$

The final act is the dot product:
$$\vec{CB} \cdot \vec{CA} = (2)(4) + (4)(3) + (8)(5) = 8 + 12 + 40 = 60$$

Through the power of vector geometry and algebraic persistence, we have arrived at the final answer of **60**.

Select Answer:

Visualized Solution

Visualizing the Triangle $A B C$

Applying Triangle Law of Vector Addition

Substituting Vector Components

Finding $α, β, γ$ in terms of $δ$

The Area of Triangle Formula

Setting up the Cross Product

Calculating the Cross Product Components

Magnitude Equation for Area

Expanding the Algebraic Terms

Simplifying into a Quadratic Equation

Solving for $δ$

Final Vectors $C B$ and $C A$

Calculating the Dot Product

Final Answer

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Triangle Law

The Area Constraint

The Algebraic Grind

Final Calculation

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