Consider a cube of side length a and whose adjacent edges are along positive x, y,z-axes. Which of the following are correct.

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

The distance of diagonal of a cube not passing through origin from origin is $a \sqrt{\frac{2}{3}}$

Angles between diagonals of cube is $\cos^{- 1} (\frac{1}{3})$

The shortest distance between two faces diagonals of a cube which are skew is $\frac{a}{\sqrt{3}}$

Angles between diagonals of cube and diagonal of face of a cube $\cos^{- 1} (\frac{2}{3})$ is

answer is A, C, D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$Drs of \bar{O G} = ⟨ a, a, a ⟩$

$Drs of \bar{B D} = ⟨ a, - a, a ⟩$

$Dr s of \bar{G A} = ⟨ 0, a, a ⟩$

$Equation of Diagonal F A = \frac{x - a}{- a} = \frac{y - a}{a} = \frac{z - a}{a}$

Any point on FA

$(- a t, a + a t, a + a t) (O M ⊥ F A)$

$a^{2} t + a^{2} + d t + a^{2} + a^{2} t = 0$

$\Rightarrow 3 a^{2} t + 2 a^{2} = 0 t = \frac{- 2}{3}$

$M (\frac{2}{3} a, \frac{1}{3} a, \frac{1}{3} a) O M^{2} = \frac{4}{9} a^{2} + \frac{1}{9} a^{2} + \frac{1}{9} a^{2}$

$O M = \sqrt{\frac{2}{3} a}$

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE