JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Three Dimensional Geometry

Let Q be the cube with the set of vertices

{(x_{1}, x_{2}, x_{3}) \in R^{3} : x_{1}, x_{2}, x_{3} {0, 1}}

. Let F be the set of all twelve lines containing the diagonals of the six faces of the cube Q. Let S be the set of all four lines containing the main diagonals of the cube Q; for instance, the line passing through the vertices (0, 0,0) and

(1, 1, 1)

is in S. For lines

ℓ_{1}

and

ℓ_{2}

, let

d (ℓ_{1}, ℓ_{2})

denote the shortest distance between them. Then the maximum value of

d (ℓ_{1}, ℓ_{2})

, as

ℓ_{1}

varies over F and

ℓ_{2}

varies over S, is

(A) $\frac{1}{6}$ ✓ Correct
(B) $\frac{1}{8}$
(C) $\frac{1}{3}$
(D) $\frac{1}{12}$

Answer: (A)

Solution

Consider the main diagonal OG from

(0, 0, 0)

(1, 1, 1)

with direction ratios

(1, 1, 1)

.

Consider a face diagonal AB from

(1, 0, 0)

(1, 1, 1)

with direction ratios

(0, 1, 1)

(or equivalently

(1, - 1, 0)

for edge from

(1, 0, 0)

(0, 1, 0)

).

For line OG:

\frac{x}{1} = \frac{y}{1} = \frac{z}{1}

For line AB (say from

(1, 0, 0)

with direction

(1, - 1, 0)

\frac{x - 1}{1} = \frac{y}{- 1} = \frac{z}{0}

The normal to both lines is:

n = \hat{i} 11 \hat{j} 1 - 1 \hat{k} 10 = \hat{i} + \hat{j} - 2 \hat{k}

The shortest distance is:

d = \frac{∣ O A \cdot n ∣}{∣ n ∣} = \frac{∣ i ^ \cdot ( i ^ + j ^ - 2 k ^ ) ∣}{1 + 1 + 4} = \frac{1}{6}

Answer: (A)

JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Single correct option

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JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Three Dimensional Geometry

Solution

More JEE Advanced questions on Three Dimensional Geometry