Question

If lengths of edges of cube are a, b, c, then the shortest distance between diagonal vector OO' and edge which is non-co-planer vector OO' to AB' is ____ 

(A) [(ac)/√(a² + c²)]

(B) [(ab)/√(a² + b²)] 

(C) [(bc)/√(b² + c²)]

(D) [(abc)/√(a² + b² + c²)]

Answer 1

The correct option (B) [(ab)/√(a² + b²)]

Explanation:

equation of vector OO' (x/a) = (y/b) = (z/c)

vector AB is non-co-planar edge to OO' then

[(x – a)/0] = (y/0) = (z/c)

∴ vector a = (0, 0, 0), ℓ = (a, b, c)

also vector b = (a, 0, 0), m = (0, 0, c)

Now vector l x vector m

Now b – a = (a, 0, 0)

vector|ℓ × m| = c√(a² + b²)

and vector (b – a) ∙ vector (ℓ × m) = (a, 0, 0) ∙ (bc, – ca, 0) = abc

shortest distance between two skew lines   

= [(abc)/{c√(a² + b²)}

= [(ab)/√(a² + b²)].