The vertices B and C of a ΔABC lie on the line, (x+2)/3 = (y-1)/0 = z/4 such that BC = 5 units. Then

Q: Why can't we just find the coordinates of B and C? Isn't that more direct?

While you could technically solve for the coordinates of $B$ and $C$ by parameterizing the line, it is computationally expensive and prone to algebraic errors. In JEE Advanced, we value elegance and efficiency. The area formula $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$ allows us to bypass finding the specific coordinates of $B$ and $C$ entirely, as the height is invariant for any segment of length 5 on that line.

Q: Why do we use the cross product to find the distance?

The cross product $\vec{AP} \times \vec{v}$ yields a vector whose magnitude represents the area of the parallelogram formed by $\vec{AP}$ and $\vec{v}$. Since the area of this parallelogram is also defined as $\text{base} \times \text{height}$ (where the base is the magnitude of the direction vector $|\vec{v}|$), dividing the magnitude of the cross product by $|\vec{v}|$ isolates the height $h$ perfectly.

Q: Does the position of the segment BC on the line matter?

That is a fantastic conceptual question! No, it does not. Because the line is straight, the perpendicular distance from a fixed point $A$ to the line is constant regardless of where the segment $BC$ is placed. As long as the length $BC$ is 5, the area remains constant. This is the beauty of 3D geometry—symmetry and invariance are your best friends.

Question

The vertices B and C of a $\Delta ABC$ lie on the line, $\frac{x+2}{3} = \frac{y-1}{0} = \frac{z}{4}$ such that $BC = 5$ units. Then the area (in sq. units) of this triangle, given that the point $A(1, -1, 2)$, is :-

Options:

(A) $2\sqrt{34}$

(B) $\sqrt{34}$

(C) $6$

(D) $5\sqrt{17}$

Sigma Percentile · Accepted Answer

### The Geometry of the Void: Solving 3D Triangles

Welcome, future engineers. Today, we are not just solving a math problem; we are navigating the vast, elegant expanse of three-dimensional space.

Imagine a triangle, $\Delta ABC$, floating in the void. Two of its vertices, $B$ and $C$, are anchored to a line $L$, while the third vertex, $A$, sits at a fixed coordinate in space.

It feels daunting, doesn't it? But remember, in the world of JEE Advanced, complexity is often just a mask for a simple, beautiful truth waiting to be uncovered.

### Phase 1: The Strategy of Invariance

We are given the base length $BC = 5$. The classic formula for the area of a triangle is:

$$	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$$

We have the base. The challenge, therefore, is entirely contained within the height $h$.

Geometrically, this height is the perpendicular distance from point $A$ to the line $L$. Notice something profound here: the area does not depend on the specific coordinates of $B$ and $C$.

As long as the distance between them is $5$, the area remains constant. We are looking for the shortest path from $A$ to the line $L$. This is our target.

### Phase 2: Decoding the Line

Look at the equation of the line:

$$\frac{x+2}{3} = \frac{y-1}{0} = \frac{z}{4}$$

This is the DNA of our line. From the numerators, we extract a point $P(-2, 1, 0)$ that lies on the line. From the denominators, we extract the direction vector $\vec{v} = 3\hat{i} + 0\hat{j} + 4\hat{k}$.

We now have a fixed point $P$ on the line and a direction vector $\vec{v}$ that defines its orientation. To find the distance from $A(1, -1, 2)$ to this line, we need to bridge the gap between $A$ and $P$.

We construct the vector $\vec{AP} = P - A$. Calculating this, we get:

$$\vec{AP} = (-2 - 1)\hat{i} + (1 - (-1))\hat{j} + (0 - 2)\hat{k} = -3\hat{i} + 2\hat{j} - 2\hat{k}$$

### Phase 3: The Power of the Cross Product

Now, we invoke the most powerful tool in our 3D arsenal: the cross product. The perpendicular distance $h$ from a point to a line is given by the elegant formula:

$$h = \frac{|\vec{AP} 	imes \vec{v}|}{|\vec{v}|}$$

Why does this work? Because the magnitude of the cross product $\vec{AP} 	imes \vec{v}$ represents the area of the parallelogram formed by these two vectors.

By dividing by the magnitude of the base vector $\vec{v}$, we are essentially calculating the height of that parallelogram, which is exactly the perpendicular distance we need.

Let us compute the cross product $\vec{AP} 	imes \vec{v}$ using the determinant method:

$$\vec{AP} 	imes \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -3 & 2 & -2 \ 3 & 0 & 4 \end{vmatrix}$$

Expanding this, we get $\hat{i}(2 	imes 4 - (-2) 	imes 0) - \hat{j}(-3 	imes 4 - (-2) 	imes 3) + \hat{k}(-3 	imes 0 - 2 	imes 3)$. This simplifies to $8\hat{i} + 6\hat{j} - 6\hat{k}$.

### Phase 4: The Final Convergence

We are almost there. We need the magnitudes. The magnitude of our cross product vector is:

$$|\vec{AP} 	imes \vec{v}| = \sqrt{8^2 + 6^2 + (-6)^2} = \sqrt{64 + 36 + 36} = \sqrt{136} = 2\sqrt{34}$$

The magnitude of the direction vector is:

$$|\vec{v}| = \sqrt{3^2 + 0^2 + 4^2} = \sqrt{9 + 0 + 16} = 5$$

Substituting these into our height formula, we get $h = \frac{2\sqrt{34}}{5}$.

Finally, we return to our area formula:

$$	ext{Area} = \frac{1}{2} 	imes 5 	imes \frac{2\sqrt{34}}{5}$$

Watch the magic happen—the $5$s cancel, the $2$s cancel, and we are left with **$\sqrt{34}$**. It is clean, it is precise, and it is correct. You have successfully navigated the 3D geometry of this problem.

Animated Solution for Mathematics - Three Dimensional Geometry: The vertices B and C of a $Δ A B C$ lie on the line, $\frac{x + 2}{3} = \frac{y - 1}{0} = \frac{z}{4}$ such that $B C = 5$ units. Then the area (in sq. units) of this triangle, given that the point $A (1, - 1, 2)$ , is :-

Select Answer:

Visualized Solution

Visualizing the 3D Geometry

The Area Strategy

Extracting Line Parameters

Defining Vector $A P$

Calculating Vector $A P$

The Perpendicular Distance Formula

Setting up the Cross Product

Evaluating the Cross Product

Magnitude of the Cross Product

Magnitude of Direction Vector

Calculating the Height $h$

Final Area Calculation

Conclusion

The Sigma Insight: Equation of a Line in Space

The Geometry of the Void

Phase 1

Phase 2

Phase 3

Phase 4

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