JEE Advanced 2024 · Paper 1 · Mathematics Q7

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MathematicsMultiple correct+4 / -2Question 7 of 51

Let $\mathbb{R}^3$ denote the three-dimensional space. Take two points $P = (1,2,3)$ and $Q = (4,2,7)$ . Let $dist(X,Y)$ denote the distance between two points $X$ and $Y$ in $\mathbb{R}^3$ . Let

$S = \left\{X \in \mathbb{R}^3 : \left(dist(X,P)\right)^2 - \left(dist(X,Q)\right)^2 = 50\right\} \text{ and}$
$T = \left\{Y \in \mathbb{R}^3 : \left(dist(Y,Q)\right)^2 - \left(dist(Y,P)\right)^2 = 50\right\}.$

Then which of the following statements is (are) TRUE?

There is a triangle whose area is 1 and all of whose vertices are from

S

There are two distinct points

L

and

M

T

such that each point on the line segment

LM

is also in

T

There are infinitely many rectangles of perimeter 48, two of whose vertices are from

S

and the other two vertices are from

T

There is a square of perimeter 48, two of whose vertices are from

S

and the other two vertices are from

T

Answer and worked solution

Official answer: A, B, C, D

Both sets are planes. For $X = (x,y,z)$ :
$|X-P|^2 - |X-Q|^2 = 2X\cdot(Q-P) + |P|^2 - |Q|^2,$
with $Q - P = (3,0,4)$ , $|P|^2 = 14$ , $|Q|^2 = 69$ .

$S$ : $2(3x+4z) - 55 = 50 \implies 6x + 8z = 105$ .
$T$ : $-(6x+8z) + 55 = 50 \implies 6x + 8z = 5$ .

They are parallel planes at distance $\dfrac{|105-5|}{\sqrt{36+64}} = 10$ .

(A) A plane contains triangles of any area. True.

(B) A plane contains whole segments. True.

(C) Put two adjacent vertices in $S$ and the other two in $T$ . The two sides crossing between the planes have length $\ell \ge 10$ (any $\ell \ge 10$ is possible by tilting), and the sides lying in the planes have length $24 - \ell$ . Every $\ell \in [10, 24)$ works, so there are infinitely many. True.

(D) Side $12 \ge 10$ , so a tilted square fits. True.

Topics:Equation of a plane Angles and distances

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JEE Advanced 2024 · Paper 1, Mathematics question 7