Let the co-ordinates of one vertex of ΔABC be A(0, 2, α) and the other two vertices lie on the line

Q: Why can't we just find the coordinates of $B$ and $C$ directly?

Because $B$ and $C$ are just points on a line; they aren't fixed. We only know the length of the segment $BC$. The triangle's area depends on the height, not the specific location of $B$ and $C$ on the line. Trying to find their coordinates would lead to an infinite number of solutions, which is not what we want.

Q: Why do we use the cross product formula for distance?

It is the most elegant way to handle 3D geometry. It avoids the mess of finding the foot of the perpendicular and then using the distance formula. It is a shortcut that saves time and reduces errors. In JEE, efficiency is just as important as accuracy.

Q: What if I get two values for $\alpha$?

That is exactly what happens! The problem gives you a constraint: $\alpha \in \mathbb{Z}$. This is a classic JEE filter. Always check your final answers against the constraints given in the question. If one value is a fraction and the other is an integer, the integer is your winner.

Question

Let the co-ordinates of one vertex of $\Delta ABC$ be $A(0, 2, \alpha)$ and the other two vertices lie on the line $\frac{x+\alpha}{5} = \frac{y-1}{2} = \frac{z+4}{3}$. For $\alpha \in \mathbb{Z}$, if the area of $\Delta ABC$ is 21 sq. units and the line segment $BC$ has length $2\sqrt{21}$ units, then $\alpha^2$ is equal to ______.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, my dear student. Today, we are not just solving a math problem; we are sculpting a 3D reality in our minds. Imagine you are standing in a vast, empty room with a fixed point $A$ floating at $(0, 2, \alpha)$.

Below you, there is a straight line $L$ defined by the equation:

$$\frac{x+\alpha}{5} = \frac{y-1}{2} = \frac{z+4}{3}$$

Two points, $B$ and $C$, rest on this line, forming a triangle $\Delta ABC$ with $A$. We are given that the area of this triangle is $21$ square units and the base $BC$ is $2\sqrt{21}$ units.

Because the area of a triangle is fundamentally tied to its height, we use the formula $Area = \frac{1}{2} 	imes base 	imes height$:

$$21 = \frac{1}{2} 	imes (2\sqrt{21}) 	imes h$$

The $2$ and the $\frac{1}{2}$ cancel out, leaving $21 = \sqrt{21} 	imes h$. Thus, the perpendicular distance $h$ from point $A$ to the line $L$ is exactly $\sqrt{21}$.

### The Power of the Cross Product

To find the distance from a point to a line in 3D space, we utilize the cross product. The perpendicular distance $h$ from a point $A$ to a line passing through point $P$ with direction vector $\vec{b}$ is given by:

$$h = \frac{|\vec{PA} 	imes \vec{b}|}{|\vec{b}|}$$

From the line equation, the point $P$ is $(-\alpha, 1, -4)$ and the direction vector is $\vec{b} = 5\hat{i} + 2\hat{j} + 3\hat{k}$. The magnitude of $\vec{b}$ is:

$$|\vec{b}| = \sqrt{5^2 + 2^2 + 3^2} = \sqrt{38}$$

Next, we construct the vector $\vec{PA}$ by subtracting the coordinates of $P$ from $A$:

$$\vec{PA} = (0 - (-\alpha))\hat{i} + (2 - 1)\hat{j} + (\alpha - (-4))\hat{k} = \alpha\hat{i} + 1\hat{j} + (\alpha + 4)\hat{k}$$

### The Algebraic Grind

We calculate the cross product $\vec{PA} 	imes \vec{b}$ using the determinant:

$$\vec{PA} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ \alpha & 1 & \alpha+4 \ 5 & 2 & 3 \end{vmatrix}$$

Expanding this determinant, we obtain:

$$\vec{PA} 	imes \vec{b} = \hat{i}(3 - 2(\alpha+4)) - \hat{j}(3\alpha - 5(\alpha+4)) + \hat{k}(2\alpha - 5)$$

Simplifying the components, we get:

$$\vec{PA} 	imes \vec{b} = (-2\alpha - 5)\hat{i} + (2\alpha + 20)\hat{j} + (2\alpha - 5)\hat{k}$$

Given $h = \sqrt{21}$, we substitute into the distance formula:

$$\sqrt{21} = \frac{|\vec{PA} 	imes \vec{b}|}{\sqrt{38}} \Rightarrow |\vec{PA} 	imes \vec{b}| = \sqrt{798}$$

Squaring both sides yields $|\vec{PA} 	imes \vec{b}|^2 = 798$. Expanding the magnitude squared:

$$(-2\alpha - 5)^2 + (2\alpha + 20)^2 + (2\alpha - 5)^2 = 798$$

Expanding the squares:

$$(4\alpha^2 + 20\alpha + 25) + (4\alpha^2 + 80\alpha + 400) + (4\alpha^2 - 20\alpha + 25) = 798$$

The $20\alpha$ terms cancel out, leaving:

$$12\alpha^2 + 80\alpha + 450 = 798 \Rightarrow 12\alpha^2 + 80\alpha - 348 = 0$$

Dividing by $4$, we arrive at the quadratic equation:

$$3\alpha^2 + 20\alpha - 87 = 0$$

### Final Calculation

We solve the quadratic by splitting the middle term:

$$3\alpha^2 + 29\alpha - 9\alpha - 87 = 0 \Rightarrow (\alpha - 3)(3\alpha + 29) = 0$$

This yields $\alpha = 3$ or $\alpha = -\frac{29}{3}$. Given the constraint that $\alpha \in \mathbb{Z}$, we reject the fraction and accept $\alpha = 3$.

The final value requested is $\alpha^2$:

$$\alpha^2 = 3^2 = \mathbf{9}$$

Enter Numerical Value:

Visualized Solution

Visualizing the Geometry

Finding the Height $h$

Identifying Line Parameters

Constructing Vector $P A$

The Distance Formula

Calculating $P A \times b$

Simplifying the Cross Product

Squaring for Simplicity

Expanding the Quadratic

Forming the Quadratic Equation

Solving for $α$

Final Answer

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Power of the Cross Product

The Algebraic Grind

Final Calculation

Similar Questions

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