JEE Main 2025 Jan 28 Shift 1, Mathematics Q6: Rectangular Coordinate System in Space

JEE Main2025Jan 28, Shift 1Mathematics

Let $A (x, y, z)$ be a point in xy-plane, which is equidistant from three points $(0, 3, 2), (2, 0, 3)$ and $(0, 0, 1)$ .

Let $B = (1, 4, - 1)$ and $C = (2, 0, - 2)$ . Then among the statements

(S1) : $△ A B C$ is an isosceles right angled triangle and
(S2) : the area of $△ A B C$ is $\frac{9 2}{2}$ .

A
both are true
B
only (S1) is true
C
only (S2) is true
D
both are false

Solution

Since $A$ lies in the xy-plane, $z = 0$ . Let $A = (x, y, 0)$ , $P = (0, 3, 2)$ , $Q = (2, 0, 3)$ , $R = (0, 0, 1)$ .

From $∣ A P ∣^{2} = ∣ A R ∣^{2}$ : $x^{2} + (y - 3)^{2} + 4 = x^{2} + y^{2} + 1$ , giving $- 6 y + 13 = 1$ , so $y = 2$ .

From $∣ A Q ∣^{2} = ∣ A R ∣^{2}$ : $(x - 2)^{2} + y^{2} + 9 = x^{2} + y^{2} + 1$ , giving $- 4 x + 13 = 1$ , so $x = 3$ .

Thus $A = (3, 2, 0)$ . With $B = (1, 4, - 1)$ , $C = (2, 0, - 2)$ :

$∣ A B ∣^{2} = 4 + 4 + 1 = 9$ , $∣ A C ∣^{2} = 1 + 4 + 4 = 9$ , $∣ B C ∣^{2} = 1 + 16 + 1 = 18$ .

So $∣ A B ∣ = ∣ A C ∣ = 3$ (isosceles) and $∣ A B ∣^{2} + ∣ A C ∣^{2} = ∣ B C ∣^{2}$ (right-angled at $A$ ). $△ A B C$ is an isosceles right triangle. S1 is true.

Area $= \frac{1}{2} ∣ A B ∣ \cdot ∣ A C ∣ = \frac{1}{2} (3) (3) = \frac{9}{2}$ , not $\frac{9 2}{2}$ . S2 is false.

$∴$ Only S1 is true.

Concept behind this question

Rectangular Coordinate System in SpaceNotes, formulas and examples →

Practice more Mathematics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →