Let A( veca) and B( vecb) be points on two skew line vecr= veca+ vecλ and vecr= vecb+u vecq and the shortest distance between the skew line is 1 , where vecp and vecq are unit vectors forming adjacent sides of a parallelogram enclosing an area of (1/2) units. If an angle between A B and the line of shortest distance is 60°, then A B=

Question

Let A( veca) and B( vecb) be points on two skew line vecr= veca+ vecλ and vecr= vecb+u vecq and the shortest distance between the skew line is 1 , where vecp and vecq are unit vectors forming adjacent sides of a parallelogram enclosing an area of (1/2) units. If an angle between A B and the line of shortest distance is 60°, then A B= (A) (1/2) (B) 2 (C) 1 (D) λ ∈ R= 0

Tardigrade Admin · Accepted Answer

1=|( vecb- veca) ⋅ (( vecp × vecq)/| vecp × vecq|)| ⇒ | veca- vecb| cos 60°=1 ⇒ A B=2

$\frac{1}{2}$

2

1

$λ \in R = {0}$

$1 = ∣ ∣ (b - a) \cdot \frac{( p \times q )}{∣ p \times q ∣} ∣ ∣$
$\Rightarrow ∣ a - b ∣ cos 6 0^{\circ} = 1$
$\Rightarrow A B = 2$

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2

1

λ∈R={0}

1=∣∣​(b−a)⋅∣p​×q​∣(p​×q​)​∣∣​ ⇒∣a−b∣cos60∘=1 ⇒AB=2

$\frac{1}{2}$

$λ \in R = {0}$

$1 = ∣ ∣ (b - a) \cdot \frac{( p \times q )}{∣ p \times q ∣} ∣ ∣$
$\Rightarrow ∣ a - b ∣ cos 6 0^{\circ} = 1$
$\Rightarrow A B = 2$