The minimum distance of origin from the locus of point satisfying the equation vec(r) xx vec(a) to =vec(b), vec(b) != 0 is (A) (|vec(a) * vec(b)|)/(a^2) (B) |(vec(a)+vec(a) xx vec(b))/(a^2)| (C) |(vec(a)-vec(a) xx vec(b))/(a^2)| (D) (|vec(a) xx vec(b)|)/(a^2)

Owen Mathis

Owen Mathis

Answered question

2022-11-11

The minimum distance of origin from the locus of point satisfying the equation r × a = b b 0 is
(A) | a . b | a 2
(B) | a + a × b a 2 |
(C) | a a × b a 2 |
(D) | a × b | a 2
My approach is as follow
r × a = b a × ( r × a ) = a × b ( a . a ) r ( a . r ) a = a × b
How will I proceed from here and what condition I need to use so that the locus of point is minimum.

Answer & Explanation

Kayleigh Cross

Kayleigh Cross

Beginner2022-11-12Added 19 answers

Note the fact that b is perpendicular to both a and r . Also, | a × r | = a r sin θ a , r , and therefore, for r to be the shortest possible, we should set θ a , r = π / 2. All in all, all the three vectors are perpendicular and we have r = b / a, and (note that | a × b | = a b) the correct answer is (D).
Zackary Diaz

Zackary Diaz

Beginner2022-11-13Added 4 answers

Break r into two components; a multiple of a and another vector perpendicular to a and b .
r = m a + n a × b
Substitute to the initial equation to obtain
n a × b × a = b n = 1 | a | 2
Finally we have
r = m a + 1 | a | 2 a × b
| r | is minimum if m = 0

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