How do you find the coordinates of B from the line equation?

Equate the line ratios to a constant and use the given x-coordinate (4) to solve for y and z.

What is the distance between B and C?

The distance BC is given as 10 units in the problem statement.

How is the height of the triangle calculated?

The height is the perpendicular distance from vertex A to the line containing BC.

What is the formula for the area of a triangle in 3D?

Area = 1/2 * base * height, where base is the length BC.

How do you calculate the perpendicular distance from a point to a line?

Use the vector formula: |(A-B) × d| / |d|, where d is the direction vector of the line.

What is the direction vector of the given line?

The direction vector is (1, 2, 3), derived from the denominators of the symmetric form.

What are the coordinates of B?

By solving the line equation for x=4, B is found to be (4, 9, 14).

What is the vector AB?

Vector AB = (4-1)i + (9-6)j + (14-3)k = 3i + 3j + 11k.

What is the result of the cross product between AB and the direction vector?

The cross product (3, 3, 11) × (1, 2, 3) results in the vector (-13, 2, 3).

What is the final area?

The area simplifies to 5√13 square units.

The vertices B and C of a triangle ABC lie on the line x/1 = (1-y)/-2 = (z-2)/3. Find the area of ΔABC

The vertices B and C of a triangle ABC lie on the line x/1 = (1-y)/-2 = (z-2)/3. Find the area of ΔABC | JEE Main Mathematics

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Math 3D Geometry

Q MCQ Triangle

The vertices B and C of a triangle ABC lie on the line $\frac{x}{1} = \frac{1-y}{-2} = \frac{z-2}{3}$. The coordinates of A and B are $(1, 6, 3)$ and $(4, 9, \alpha)$ respectively and C is at a distance of 10 units from B. The area (in sq. units) of $\triangle ABC$ is :

A) $20\sqrt{13}$ B) $5\sqrt{13}$ C) $15\sqrt{13}$ D) $10\sqrt{13}$

✅ Correct Answer

5√13

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Solution Steps

Standardize the Line Equation

The given line is $\frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3}$.

The direction vector of the line is $\vec{d} = \hat{i} + 2\hat{j} + 3\hat{k}$.

Find Coordinates of B and the Value of $\alpha$

B $(4, 9, \alpha)$ lies on the line. Plugging $x=4$ into the line equation:

$\frac{4}{1} = \frac{9-1}{2} = \frac{\alpha-2}{3} \implies 4 = 4 = \frac{\alpha-2}{3}$.

$\alpha – 2 = 12 \implies \alpha = 14$. Thus, $B = (4, 9, 14)$.

Calculate the Height of the Triangle

The height $h$ is the perpendicular distance from $A(1, 6, 3)$ to the line through $B$.

Let $\vec{AB} = (4-1)\hat{i} + (9-6)\hat{j} + (14-3)\hat{k} = 3\hat{i} + 3\hat{j} + 11\hat{k}$.

$h = \frac{|\vec{AB} \times \vec{d}|}{|\vec{d}|}$.

$\vec{AB} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 3 & 11 \\ 1 & 2 & 3 \end{vmatrix} = \hat{i}(9-22) – \hat{j}(9-11) + \hat{k}(6-3) = -13\hat{i} + 2\hat{j} + 3\hat{k}$.

$|\vec{AB} \times \vec{d}| = \sqrt{(-13)^2 + 2^2 + 3^2} = \sqrt{169 + 4 + 9} = \sqrt{182}$.

$|\vec{d}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}$.

$h = \sqrt{\frac{182}{14}} = \sqrt{13}$.

Calculate the Area

Base $BC = 10$ (given).

Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times \sqrt{13} = 5\sqrt{13}$.

Area = 5√13 (Option B)

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Theory

1. Perpendicular Distance Point to Line (Vector)

In 3D geometry, the shortest distance from a point $P$ to a line passing through point $B$ with direction vector $\vec{d}$ is given by $h = \frac{|(\vec{P}-\vec{B}) \times \vec{d}|}{|\vec{d}|}$. This formula uses the magnitude of the cross product to find the area of the parallelogram formed by vectors $\vec{PB}$ and $\vec{d}$, then divides by the base length $|\vec{d}|$ to isolate the altitude.

2. Symmetric Form of a Line

The symmetric form of a line equation is $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$. In this problem, we must be careful with signs, such as converting $(1-y)/-2$ to $(y-1)/2$ to correctly identify the direction ratios $(a, b, c)$. Direction ratios represent the components of a vector parallel to the line, which is the most critical element for calculating intersections or distances in 3D space.

3. Area of Triangle in 3D

While the standard area formula $\frac{1}{2}bh$ works perfectly if the altitude is known, one can also use the vector cross product directly if two side vectors are known: Area $= \frac{1}{2} |\vec{AB} \times \vec{AC}|$. In this specific problem, since the base length $BC$ was explicitly provided, the $1/2 \times \text{base} \times \text{height}$ approach was computationally more efficient.

4. Condition for a Point to Lie on a Line

A point $(x_0, y_0, z_0)$ lies on a line if its coordinates satisfy the ratios of the line equation. This property is frequently used in JEE to find unknown coordinates or parameters (like $\alpha$ in this question) by substituting one known coordinate into the line’s ratio to solve for the others.

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FAQs

Why did $\frac{1-y}{-2}$ become $\frac{y-1}{2}$?

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To match the standard form $\frac{y-y_1}{b}$, we multiply both the numerator and denominator of $\frac{1-y}{-2}$ by -1. This ensures the coefficient of $y$ is $+1$.

Could we find point C first?

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You could, using the distance formula and the line’s parametric form, but it’s unnecessary since we only need the length $BC$, which is already given as 10.

Is the distance 10 units used for height?

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No, 10 units is the length of the base $BC$. The height must be perpendicular to this base, starting from vertex $A$.

What if vertex A also lied on the line?

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If A lied on the line, the points would be collinear, the height would be 0, and the area of the triangle would be 0.

Is there a determinant formula for area in 3D?

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Yes, the area is $\frac{1}{2}\sqrt{\Delta_x^2 + \Delta_y^2 + \Delta_z^2}$, where $\Delta$ are determinants formed by coordinate pairs, but the cross-product method is essentially the same thing.

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