Let three vectors a = αi + 4j + 2k, b = 5i + 3j + 4k, c = xi + yj + zk form a triangle such that c =

Q: Wait, why do we use the cross product? Can't we just use the dot product to find the area?

Great question! The dot product $\vec{a} \cdot \vec{b}$ gives you information about the angle between vectors (via $\cos \theta$), which is useful for projections or work done. However, the area of a triangle is inherently tied to the 'perpendicular' component of the vectors, which is exactly what the cross product $\vec{a} \times \vec{b}$ captures. The cross product magnitude gives the area of the parallelogram, and we simply halve it.

Q: Why do we reject the solution $\alpha = 0$?

Always keep an eye on the problem constraints! The problem explicitly states that $\alpha$ is a positive real number. In algebra, we often find multiple roots, but in physics and geometry, we must filter them through the 'reality check' of the problem statement. Since $\alpha = 0$ would violate the condition, we discard it.

Q: Does the order of subtraction in $\vec{c} = \vec{a} - \vec{b}$ matter for the final answer?

It matters for the direction of the vector $\vec{c}$, but notice that the question asks for $|\vec{c}|^2$. Since the square of a magnitude is the sum of the squares of the components, the sign of the components doesn't change the final result. Whether you calculate $\vec{a} - \vec{b}$ or $\vec{b} - \vec{a}$, the magnitude squared remains identical.

Question

Let three vectors $\vec{a} = \alpha\hat{i} + 4\hat{j} + 2\hat{k}, \vec{b} = 5\hat{i} + 3\hat{j} + 4\hat{k}, \vec{c} = x\hat{i} + y\hat{j} + z\hat{k}$ form a triangle such that $\vec{c} = \vec{a} - \vec{b}$ and the area of the triangle is $5\sqrt{6}$. If $\alpha$ is a positive real number, then $|\vec{c}|^2$ is equal to:

Options:

(A) 16

(B) 14

(C) 12

(D) 10

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional coordinate system. You are holding two vectors, $\vec{a}$ and $\vec{b}$, which originate from the same point.

Vector $\vec{a} = \alpha\hat{i} + 4\hat{j} + 2\hat{k}$ contains an unknown component $\alpha$, while $\vec{b} = 5\hat{i} + 3\hat{j} + 4\hat{k}$ is fully defined.

These two vectors form the sides of a triangle. When we define the third side as $\vec{c} = \vec{a} - \vec{b}$, we are essentially drawing a line from the tip of $\vec{b}$ to the tip of $\vec{a}$.

### The Bridge to Area

We are given that the area of this triangle is $5\sqrt{6}$. To bridge the gap between vector components and physical area, we utilize the cross product.

The magnitude of the cross product, $|\vec{a} 	imes \vec{b}|$, represents the area of the parallelogram formed by the two vectors. Since our triangle is exactly half of that parallelogram, we use the following relation:

$$	ext{Area} = \frac{1}{2} |\vec{a} 	imes \vec{b}| = 5\sqrt{6}$$

Multiplying both sides by $2$, we obtain our target magnitude:

$$|\vec{a} 	imes \vec{b}| = 10\sqrt{6}$$

### The Determinant Dance

To calculate $\vec{a} 	imes \vec{b}$, we set up the standard $3 	imes 3$ determinant:

$$\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ \alpha & 4 & 2 \ 5 & 3 & 4 \end{vmatrix}$$

Expanding this along the first row, we compute the components:

- For $\hat{i}$: $(4 \cdot 4) - (3 \cdot 2) = 16 - 6 = 10$
- For $\hat{j}$: $-(\alpha \cdot 4 - 5 \cdot 2) = -(4\alpha - 10)$
- For $\hat{k}$: $(\alpha \cdot 3 - 5 \cdot 4) = 3\alpha - 20$

Thus, the cross product vector is $10\hat{i} - (4\alpha - 10)\hat{j} + (3\alpha - 20)\hat{k}$.

### The Algebraic Resolution

We know the magnitude of this vector is $10\sqrt{6}$. Squaring both sides to simplify the calculation, we get:

$$|\vec{a} 	imes \vec{b}|^2 = (10\sqrt{6})^2 = 600$$

Substituting our components into the magnitude formula, we have:

$$10^2 + (4\alpha - 10)^2 + (3\alpha - 20)^2 = 600$$

Expanding the squares yields:

$$100 + (16\alpha^2 - 80\alpha + 100) + (9\alpha^2 - 120\alpha + 400) = 600$$

Grouping the terms results in $25\alpha^2 - 200\alpha + 600 = 600$. The constants cancel out, leaving:

$$25\alpha^2 - 200\alpha = 0 \implies 25\alpha(\alpha - 8) = 0$$

Given the constraint $\alpha > 0$, we reject $\alpha = 0$ and conclude that **$\alpha = 8$**.

### The Final Destination

With $\alpha = 8$, our vector $\vec{a}$ is $8\hat{i} + 4\hat{j} + 2\hat{k}$. We now determine $\vec{c} = \vec{a} - \vec{b}$:

$$\vec{c} = (8-5)\hat{i} + (4-3)\hat{j} + (2-4)\hat{k} = 3\hat{i} + \hat{j} - 2\hat{k}$$

Finally, we calculate $|\vec{c}|^2$ by summing the squares of the components:

$$|\vec{c}|^2 = 3^2 + 1^2 + (-2)^2 = 9 + 1 + 4 = 14$$

The final answer is **$14$**.

Select Answer:

Visualized Solution

Visualizing the Vectors

The Third Side $c$

Area of the Triangle

Magnitude of Cross Product

Setting up the Cross Product

Expanding the Determinant

Squaring the Magnitude

Algebraic Expansion

Solving for $α$

Finding Vector $c$

Final Calculation: $∣ c ∣^{2}$

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Bridge to Area

The Determinant Dance

The Algebraic Resolution

The Final Destination

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