Let in a △ ABC the length of the side AC be 6, the vertex B be (1, 2, 3) and the vertices A, C lie o

Q: Wait, do I need to find the coordinates of points A and C to solve this?

That is a common trap! You might be tempted to solve for A and C, but notice that the problem doesn't provide enough constraints to fix them. The beauty of this problem is that the height is invariant regardless of where A and C are on the line, as long as the distance $AC$ is fixed. Focus on the geometry, not the coordinates!

Q: Why is the cross product formula used for distance? Can't I just use projection?

You absolutely can use projection, but the cross product method is often faster and less prone to sign errors. It essentially calculates the area of the parallelogram formed by $\vec{BD}$ and $\vec{d}$ and divides by the base length $|\vec{d}|$. It is a powerful tool in your JEE arsenal—keep it sharp!

Q: I got $\sqrt{833/17}$ and panicked. Did I do something wrong?

Take a deep breath! In JEE Advanced, if you see a messy fraction under a square root, it is almost always a perfect square waiting to be revealed. $833 \div 17$ is exactly $49$. Trust your process, and the numbers will often simplify beautifully at the end.

Question

Let in a $	riangle ABC$ the length of the side AC be 6, the vertex B be (1, 2, 3) and the vertices A, C lie on the line $\frac{x-6}{3}=rac{y-7}{2}=rac{z-7}{-2}.$ Then the area (in sq. units) of $	riangle ABC$ is:

Options:

(A) 17

(B) 21

(C) 56

(D) 42

Sigma Percentile · Accepted Answer

### Analyzing the Geometry of the Void

Imagine you are standing in a vast, three-dimensional coordinate space. You have a fixed point $B$ at $(1, 2, 3)$ and a line $L$ stretching out into the distance. Two other points, $A$ and $C$, are dancing along this line.

You are asked to find the area of the triangle formed by these three points. The beauty of geometry lies in finding the invariants. No matter where $A$ and $C$ slide along that line, as long as the distance between them remains $6$, the area of the triangle remains constant.

This occurs because the height of the triangle—the perpendicular distance from $B$ to the line—never changes.

### The Strategy: Deconstructing the Problem

We know the fundamental formula for the area of a triangle:
$$	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$$

We are given the base $AC = 6$. Our mission is clear: find the height $h$. In 3D geometry, the height is the perpendicular distance from a point to a line.

We do not need to find the coordinates of $A$ or $C$. We simply need the distance from $B$ to the line $L$.

### The Vector Toolkit

To find this distance, we use the elegant power of vectors. If we have a point $D$ on the line and a direction vector $\vec{d}$ of the line, the perpendicular distance $h$ is given by the formula:

$$h = \frac{|\vec{BD} 	imes \vec{d}|}{|\vec{d}|}$$

This formula is a masterpiece of efficiency. It calculates the area of the parallelogram formed by the vector $\vec{BD}$ and the direction vector $\vec{d}$, and then divides by the length of the base $\vec{d}$ to isolate the height.

From the line equation $\frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2}$, we extract the point $D = (6, 7, 7)$ and the direction vector $\vec{d} = 3\hat{i} + 2\hat{j} - 2\hat{k}$.

### The Calculation: Precision and Patience

Now, we construct the vector $\vec{BD} = \vec{D} - \vec{B} = (6-1)\hat{i} + (7-2)\hat{j} + (7-3)\hat{k} = 5\hat{i} + 5\hat{j} + 4\hat{k}$. Next, we compute the cross product $\vec{BD} 	imes \vec{d}$ using the determinant method:

$$\vec{BD} 	imes \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 5 & 5 & 4 \ 3 & 2 & -2 \end{vmatrix}$$

Expanding this, we get $\hat{i}(-10 - 8) - \hat{j}(-10 - 12) + \hat{k}(10 - 15)$, which simplifies to $-18\hat{i} + 22\hat{j} - 5\hat{k}$.

The magnitude of this vector is:
$$\sqrt{(-18)^2 + 22^2 + (-5)^2} = \sqrt{324 + 484 + 25} = \sqrt{833}$$

The magnitude of the direction vector $\vec{d}$ is:
$$\sqrt{3^2 + 2^2 + (-2)^2} = \sqrt{9 + 4 + 4} = \sqrt{17}$$

### The Grand Finale

Putting it all together, the height $h$ is calculated as:
$$h = \frac{\sqrt{833}}{\sqrt{17}} = \sqrt{\frac{833}{17}} = \sqrt{49} = 7$$

The height is exactly $7$. Now, we return to our area formula:
$$	ext{Area} = \frac{1}{2} 	imes 6 	imes 7 = 21$$

We have conquered the 3D space. Remember, in JEE Advanced, the complexity is often a mask for a simple, elegant geometric truth. Keep your vectors sharp and your logic clear. The final answer is **21**.

Animated Solution for Mathematics - Three Dimensional Geometry: Let in a $△ A B C$ the length of the side AC be 6, the vertex B be (1, 2, 3) and the vertices A, C lie on the line $\frac{x-6}{3}= rac{y-7}{2}= rac{z-7}{-2}.$ Then the area (in sq. units) of $△ A B C$ is:

Select Answer:

Visualized Solution

Visualizing the 3D Triangle

The Area Formula

Defining the Height $h$

Perpendicular Distance Formula

Extracting Line Parameters

Constructing Vector $B D$

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

The Sigma Insight: Equation of a Line in Space

Analyzing the Geometry of the Void

The Strategy

The Vector Toolkit

The Calculation

The Grand Finale

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