Let A be the point of intersection of the lines L1: (x-7)/1 = (y-5)/0 = (z-3)/-1 and L2: (x-1)/3 = (

Q: Wait, do I really need to find the coordinates of point $A$?

Great question! In this specific problem, the area formula $\frac{1}{2} AB \cdot AC \sin \theta$ only depends on the lengths $AB$, $AC$, and the angle $\theta$. Since $AB$ and $AC$ are given as constants, you technically don't need the coordinates of $A$ to calculate the area. However, finding $A$ is a great sanity check to ensure the lines actually intersect!

Q: Why is the cosine of the angle negative? Did I calculate the dot product wrong?

Not at all! A negative cosine simply means the angle between the direction vectors is obtuse (greater than $90^\circ$). In geometry, the angle between two lines is usually taken as the acute angle, but for the area formula, $\sin \theta$ is always positive, so the sign of $\cos \theta$ doesn't affect the final area. You are on the right track!

Q: Can I use the cross product to find the area?

Absolutely! The area of a triangle formed by vectors $\vec{AB}$ and $\vec{AC}$ is $\frac{1}{2} |\vec{AB} \times \vec{AC}|$. However, since you don't have the explicit vectors $\vec{AB}$ and $\vec{AC}$ (only their magnitudes), the $\frac{1}{2}ab \sin \theta$ approach is much faster and less prone to calculation errors. Stick to the trigonometric method for efficiency!

Question

Let $A$ be the point of intersection of the lines $L_1: \frac{x-7}{1} = \frac{y-5}{0} = \frac{z-3}{-1}$ and $L_2: \frac{x-1}{3} = \frac{y+3}{4} = \frac{z+7}{5}$. Let $B$ and $C$ be the point on the lines $L_1$ and $L_2$ respectively such that $AB = AC = \sqrt{15}$. Then the square of the area of the triangle $ABC$ is :

Options:

(A) $54$

(B) $63$

(C) $57$

(D) $60$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future IITians! Today, we are going to peel back the layers of a problem that might look like a coordinate geometry nightmare but is, in reality, a beautiful, elegant dance of vectors.

Imagine you are standing in a vast 3D space. You see two lines, $L_1$ and $L_2$, slicing through the air, destined to meet at a single point, $A$. This is the heart of our problem.

### The Intersection

First, we must locate point $A$. The lines are given in their symmetric form. By equating them to parameters $\lambda$ and $\mu$, we can express any point on these lines.

At the intersection point $A$, the coordinates must be identical. By equating the $y$-coordinates, we find $\mu = 2$. Substituting this back, we find $A = (7, 5, 3)$.

It is a clean, satisfying result. But here is the secret: for the area of the triangle $ABC$, we actually only need the angle between the lines. The intersection point $A$ is our anchor, but the magic happens in the angle.

### The Trigonometric Bridge

We have an isosceles triangle $ABC$ with $AB = AC = \sqrt{15}$. The area of any triangle is given by the formula:

$$	ext{Area} = \frac{1}{2} \cdot 	ext{side}_1 \cdot 	ext{side}_2 \cdot \sin 	heta$$

We have the sides, but we need $\sin 	heta$. We look at the direction vectors of the lines to find the angle $	heta$. From the denominators of our line equations, we extract $\vec{d_1} = (1, 0, -1)$ and $\vec{d_2} = (3, 4, 5)$.

The angle $	heta$ between the lines is the angle between these vectors. We use the dot product formula:

$$\cos 	heta = \frac{\vec{d_1} \cdot \vec{d_2}}{|\vec{d_1}| |\vec{d_2}|}$$

Calculating this, we get:

$$\cos 	heta = \frac{(1)(3) + (0)(4) + (-1)(5)}{\sqrt{1^2+0^2+(-1)^2} \sqrt{3^2+4^2+5^2}} = \frac{-2}{\sqrt{2} \sqrt{50}} = \frac{-2}{10} = -\frac{1}{5}$$

Do not panic at the negative sign! It just tells us the angle is obtuse. For our area formula, we need $\sin 	heta$. Using the identity $\sin 	heta = \sqrt{1 - \cos^2 	heta}$, we find:

$$\sin 	heta = \sqrt{1 - \left(-\frac{1}{5}ight)^2} = \sqrt{\frac{24}{25}} = \frac{2\sqrt{6}}{5}$$

### The Final Victory

Now, we bring it all together. The area of $	riangle ABC$ is $\frac{1}{2} \cdot AB \cdot AC \cdot \sin 	heta$. Substituting our values:

$$	ext{Area} = \frac{1}{2} \cdot \sqrt{15} \cdot \sqrt{15} \cdot \frac{2\sqrt{6}}{5}$$

The $\sqrt{15} \cdot \sqrt{15}$ becomes $15$, and the $2$s cancel out. We are left with:

$$	ext{Area} = \frac{15 \cdot \sqrt{6}}{5} = 3\sqrt{6}$$

The question asks for the square of the area. We square $3\sqrt{6}$ to get $9 \cdot 6 = 54$.

The final result is **54**. You have mastered the geometry. Keep this momentum going!

Select Answer:

Visualized Solution

Visualizing the Lines $L_{1}$ and $L_{2}$

Parametric Representation

Solving for Intersection Point $A$

Finding Coordinates of $A$

Setting up Triangle $A B C$

Identifying Direction Vectors

Calculating $cos θ$

Evaluating $cos θ$

Calculating $sin θ$

Area Formula for $△ A B C$

Calculating the Area

Final Answer: Square of the Area

The Sigma Insight: Angle Between Two Lines

Analyzing the Setup

The Intersection

The Trigonometric Bridge

The Final Victory

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Select Answer:

Visualized Solution

Visualizing the Lines .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

Parametric Representation

Solving for Intersection Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Finding Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Setting up Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ABC