JEE Main 2024 (27 Jan Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Three Dimensional Geometry: The position vectors of the vertices $A, B$ and $C$ of a triangle are $2 \hat{i} - 3 \hat{j} + 3 \hat{k}$ , $2 \hat{i} + 2 \hat{j} + 3 \hat{k}$ and $- \hat{i} + \hat{j} + 3 \hat{k}$ respectively. Let $l$ denotes the length of the angle bisector $A D$ of $∠ B A C$ where $D$ is on the line segment $B C$ , then $2 l^{2}$ equals :

Q: Why is the triangle planar? Does this change the vector approach?

It is planar because all $z$-coordinates are $3$. This means the triangle lies entirely in the plane $z=3$. It does not change the vector approach, but it simplifies the calculation significantly because the $z$-component of the distance formula will always be zero, effectively reducing a 3D problem to a 2D one.

Q: Why can't we just use the Angle Bisector Theorem formula?

You absolutely can! The Angle Bisector Theorem states that $D$ divides $BC$ in the ratio $AB:AC$. Since our calculation shows $AB=AC=5$, the ratio is $1:1$, which means $D$ is the midpoint. The isosceles property is just a faster, more intuitive way to see this.

Select Answer:

Visualized Solution

Visualizing the Triangle

Vertices of $△ A B C$ are given as position vectors.
$A = 2 \hat{i} - 3 \hat{j} + 3 \hat{k} ⟹ (2, - 3, 3)$
$B = 2 \hat{i} + 2 \hat{j} + 3 \hat{k} ⟹ (2, 2, 3)$
$C = - \hat{i} + \hat{j} + 3 \hat{k} ⟹ (- 1, 1, 3)$
Notice all $z$ -coordinates are $3$ . The triangle lies in a plane parallel to the $x y$ -plane.

Strategy: Check Side Lengths

To understand the geometry of the triangle, we first calculate the lengths of the sides forming $∠ B A C$ .
Distance formula: $d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}$

Setting up Length $A B$

Substitute coordinates of $A (2, - 3, 3)$ and $B (2, 2, 3)$ :
$A B = (2 - 2)^{2} + (2 - (- 3))^{2} + (3 - 3)^{2}$

Calculating $A B$

$A B = 0^{2} + 5^{2} + 0^{2}$
$A B = 25 = 5$

Setting up Length $A C$

Substitute coordinates of $A (2, - 3, 3)$ and $C (- 1, 1, 3)$ :
$A C = (- 1 - 2)^{2} + (1 - (- 3))^{2} + (3 - 3)^{2}$

Calculating $A C$

$A C = (- 3)^{2} + 4^{2} + 0^{2}$
$A C = 9 + 16 = 25 = 5$

The Isosceles Property

Since $A B = A C = 5$ , $△ A B C$ is an isosceles triangle.
Theorem: In an isosceles triangle, the angle bisector of the vertex angle is also the median to the opposite side.
Therefore, $A D$ bisects $B C$ .

Midpoint Formula for $D$

Since $A D$ is the median, $D$ is the midpoint of $B C$ .
$D = (\frac{x _{B} + x _{C}}{2}, \frac{y _{B} + y _{C}}{2}, \frac{z _{B} + z _{C}}{2})$
$D = (\frac{2 + ( - 1 )}{2}, \frac{2 + 1}{2}, \frac{3 + 3}{2})$

Coordinates of $D$

$D = (\frac{1}{2}, \frac{3}{2}, 3)$

Setting up Length $A D$ ( $l$ )

We need the length $l = A D$ .
$A (2, - 3, 3)$ and $D (\frac{1}{2}, \frac{3}{2}, 3)$
$l^{2} = (\frac{1}{2} - 2)^{2} + (\frac{3}{2} - (- 3))^{2} + (3 - 3)^{2}$

Calculating $l^{2}$

$l^{2} = (- \frac{3}{2})^{2} + (\frac{9}{2})^{2} + 0$
$l^{2} = \frac{9}{4} + \frac{81}{4}$
$l^{2} = \frac{90}{4} = \frac{45}{2}$

Final Computation

The question asks for the value of $2 l^{2}$ .
$2 l^{2} = 2 \times \frac{45}{2}$
$2 l^{2} = 45$

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The Sigma Insight: Direction Cosines and Direction Ratios

Solution Diagram

Analyzing the Setup

Welcome, future engineer. Today, we are going to dissect a problem that looks like a 3D vector nightmare but is actually a beautiful geometric dance.

Let's start by looking at the vertices:

A (2, - 3, 3)

B (2, 2, 3)

, and

C (- 1, 1, 3)

Did you notice the

z

-coordinate? It is

3

for every single point. This is the first secret the problem setter hid in plain sight. The triangle is not floating in space; it is resting perfectly on the plane

z = 3

. This immediately reduces our 3D problem to a 2D one.

Calculating Side Lengths

Now, let's calculate the side lengths. Using the distance formula

d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}

we find the length of

A B

A B = (2 - 2)^{2} + (2 - (- 3))^{2} + (3 - 3)^{2} = 0^{2} + 5^{2} + 0^{2} = 5

Similarly, for

A C

, we get:

A C = (- 1 - 2)^{2} + (1 - (- 3))^{2} + (3 - 3)^{2} = (- 3)^{2} + 4^{2} + 0^{2} = 9 + 16 = 5

The Geometric Insight

This is the 'Aha!' moment. We have an isosceles triangle!

In an isosceles triangle, the angle bisector from the vertex is also the median. This means

D

is the midpoint of

B C

We find

D

by averaging the coordinates of

B

and

C

D = (\frac{2 - 1}{2}, \frac{2 + 1}{2}, \frac{3 + 3}{2}) = (\frac{1}{2}, \frac{3}{2}, 3)

Final Calculation

Finally, we calculate the length

A D

using the distance formula:

l^{2} = A D^{2} = (\frac{1}{2} - 2)^{2} + (\frac{3}{2} - (- 3))^{2} + (3 - 3)^{2}

l^{2} = (- \frac{3}{2})^{2} + (\frac{9}{2})^{2} = \frac{9}{4} + \frac{81}{4} = \frac{90}{4} = \frac{45}{2}

The question asks for

2 l^{2}

, so:

2 \times \frac{45}{2} = 45

It is elegant, it is precise, and it is pure JEE Advanced logic.

Common Traps & FAQs

Why is the triangle planar? Does this change the vector approach?

It is planar because all

z

-coordinates are

3

. This means the triangle lies entirely in the plane

z = 3

. It does not change the vector approach, but it simplifies the calculation significantly because the

z

-component of the distance formula will always be zero, effectively reducing a 3D problem to a 2D one.

Why can't we just use the Angle Bisector Theorem formula?

You absolutely can! The Angle Bisector Theorem states that

D

divides

B C

in the ratio

A B : A C

. Since our calculation shows

A B = A C = 5

, the ratio is

1 : 1

, which means

D

is the midpoint. The isosceles property is just a faster, more intuitive way to see this.

Is it always safe to assume the triangle is isosceles?

No! Never assume. Always calculate the side lengths first. We only used the isosceles property because our calculation verified that

A B = A C = 5

. In JEE Advanced, always verify your geometric assumptions with the given data.

Select Answer:

Visualized Solution

Visualizing the Triangle

Strategy: Check Side Lengths

Setting up Length $A B$

Calculating $A B$

Setting up Length $A C$

Calculating $A C$

The Isosceles Property

Midpoint Formula for $D$

Coordinates of $D$

Setting up Length $A D$ ( $l$ )

Calculating $l^{2}$

Final Computation

The Sigma Insight: Direction Cosines and Direction Ratios

Analyzing the Setup

Calculating Side Lengths

The Geometric Insight

Final Calculation

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Select Answer:

Visualized Solution

Visualizing the Triangle

Strategy: Check Side Lengths

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

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB

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

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AC

The Isosceles Property

Midpoint Formula for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D

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

Setting up Length .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AD ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } l)

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } l2

Final Computation

The Sigma Insight: Direction Cosines and Direction Ratios

Analyzing the Setup

Calculating Side Lengths

The Geometric Insight

Final Calculation

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Setting up Length $A B$

Calculating $A B$

Setting up Length $A C$

Calculating $A C$

Midpoint Formula for $D$

Coordinates of $D$

Setting up Length $A D$ ( $l$ )

Calculating $l^{2}$

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