Let a → ,   b → ,   c → be three vectors mutually perpendicular to each other…

Let a, b, c be three vectors mutually perpendicular to each other and have same magnitude. If a vector r satisfies a×{r-b×a}+b×{r-c×b}+c×{r-a×c}=0, then r is equal to:
  1. 13(a+b+c)
  2. 13(2a+b-c)
  3. 12(a+b+c)
  4. 12(a+b+2c)

Solution

Given:a×{r-b×a}+b×{r-c×b}+c×{r-a×c}=0

(a.a)(r-b)-(a·(r-b))a+(b·b)(r-c)-(b·(r-c))b+(c.c)(r-a)-(c.(r-a))c=0

|a|2(r-b)-(r·a)a+|b|2(r-c)-(r·b)b+|c|2(r-a)-(r·c)c=0

|a|2[3r-(a+b+c)]-[(r·a)a)+(r·b)b+(r·c)c]=0 |a|2=|b|2=|c|2

|a|2[3r-(a+b+c)-(r)]=0(r·a)a+(r·b)b+(r·c)c=a2r  , |a|2=|b|2=|c|2

 3r-(a+b+c)-r=0

r=a+b+c2

Asked in: JEE Main 2021 (31 Aug Shift 2)

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