AIEEE 2005

JEE Main 2005 Mathematics question on Vector Algebra

If a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} are three non-zero, non-coplanar vectors and b⃗1=b⃗−b⃗⋅a⃗∣a⃗∣2a⃗\vec{b}_1 = \vec{b} - \dfrac{\vec{b}\cdot\vec{a}}{|\vec{a}|^2}\vec{a}, b⃗2=b⃗+b⃗⋅a⃗∣a⃗∣2a⃗\vec{b}_2 = \vec{b} + \dfrac{\vec{b}\cdot\vec{a}}{|\vec{a}|^2}\vec{a},

c⃗1=c⃗−c⃗⋅a⃗∣a⃗∣2a⃗−c⃗⋅b⃗∣b⃗∣2b⃗\vec{c}_1 = \vec{c} - \dfrac{\vec{c}\cdot\vec{a}}{|\vec{a}|^2}\vec{a} - \dfrac{\vec{c}\cdot\vec{b}}{|\vec{b}|^2}\vec{b}, c⃗2=c⃗−c⃗⋅a⃗∣a⃗∣2a⃗−c⃗⋅b⃗1∣b⃗1∣2b⃗1\vec{c}_2 = \vec{c} - \dfrac{\vec{c}\cdot\vec{a}}{|\vec{a}|^2}\vec{a} - \dfrac{\vec{c}\cdot\vec{b}_1}{|\vec{b}_1|^2}\vec{b}_1,

c⃗3=c⃗−c⃗⋅a⃗∣a⃗∣2a⃗−c⃗⋅b⃗2∣b⃗2∣2b⃗2\vec{c}_3 = \vec{c} - \dfrac{\vec{c}\cdot\vec{a}}{|\vec{a}|^2}\vec{a} - \dfrac{\vec{c}\cdot\vec{b}_2}{|\vec{b}_2|^2}\vec{b}_2, c⃗4=c⃗−c⃗⋅a⃗∣a⃗∣2a⃗\vec{c}_4 = \vec{c} - \dfrac{\vec{c}\cdot\vec{a}}{|\vec{a}|^2}\vec{a}

Then, which of the following is a set of mutually orthogonal vectors?
  1. A.
    {a⃗,b⃗1,c⃗1}\{\vec{a}, \vec{b}_1, \vec{c}_1\}
  2. B.
    {a⃗,b⃗1,c⃗2}\{\vec{a}, \vec{b}_1, \vec{c}_2\}
  3. C.
    {a⃗,b⃗2,c⃗3}\{\vec{a}, \vec{b}_2, \vec{c}_3\}
  4. D.
    {a⃗,b⃗2,c⃗4}\{\vec{a}, \vec{b}_2, \vec{c}_4\}
Show answer

Correct option: B

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