If u¯, m, r¯ be three mutually perpendicular vectors with same magnitude. If e¯ satisfies the relation u¯×e¯−m¯×u¯+m¯×(e¯−r¯)×m¯+r¯×e¯−u¯×r¯=0¯ then e¯=

Since u¯,m¯,r¯ are mutually perpendicular vectors of same magnitude, we can resolve e¯ along the directions u¯,m¯,r¯ let u¯=m¯=r¯=K Let e¯=au¯+bm¯+cr¯ u¯×e¯−m¯×u¯=u¯.u¯e¯−m¯−u¯.e¯−m¯u¯=K2(e¯−m¯)−u¯.e¯u¯∵u¯.m¯=m¯.r¯=m¯.u¯=0=K2e¯−m¯−u¯.e¯u¯ Similar m¯×(e¯−r¯)×m¯=m¯.m¯e¯−r¯−m¯.e¯−r¯m¯ =K2e¯−r¯−m¯.e¯m¯r¯×e¯−u¯×r¯=r¯.r¯e¯−u¯−r¯.e¯−u¯.r¯ =K2e¯−u¯−r¯.e¯r Substitute above, in the given question e¯=12u¯+m¯+r¯

If u¯, m, r¯ be three mutually perpendicular vectors with same magnitude. If e¯ satisfies the relation u¯×e¯−m¯×u¯+m¯×(e¯−r¯)×m¯+r¯×e¯−u¯×r¯=0¯ then e¯=

Since u¯,m¯,r¯ are mutually perpendicular vectors of same magnitude, we can resolve e¯ along the directions u¯,m¯,r¯ let u¯=m¯=r¯=K Let e¯=au¯+bm¯+cr¯ u¯×e¯−m¯×u¯=u¯.u¯e¯−m¯−u¯.e¯−m¯u¯=K2(e¯−m¯)−u¯.e¯u¯∵u¯.m¯=m¯.r¯=m¯.u¯=0=K2e¯−m¯−u¯.e¯u¯ Similar m¯×(e¯−r¯)×m¯=m¯.m¯e¯−r¯−m¯.e¯−r¯m¯ =K2e¯−r¯−m¯.e¯m¯r¯×e¯−u¯×r¯=r¯.r¯e¯−u¯−r¯.e¯−u¯.r¯ =K2e¯−u¯−r¯.e¯r Substitute above, in the given question e¯=12u¯+m¯+r¯

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If $\bar{u}, m, \bar{r}$ be three mutually perpendicular vectors with same magnitude. If $\bar{e}$ satisfies the relation $\bar{u} \times \{(\bar{e} - \bar{m}) \times \bar{u}\} + \bar{m} \times \{(\bar{e} - \bar{r}) \times \bar{m}\} + \bar{r} \times \{(\bar{e} - \bar{u}) \times \bar{r}\} = \bar{0}$ then $\bar{e} =$

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$\bar{0}$

$\frac{1}{4} (\bar{u} + \bar{m} + \bar{r})$

$\frac{1}{3} (\bar{u} + \bar{m} + \bar{r})$

$\frac{1}{2} (\bar{u} + \bar{m} + \bar{r})$

answer is B.

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Detailed Solution

Since $\bar{u}, \bar{m}, \bar{r}$ are mutually perpendicular vectors of same magnitude, we can resolve $\bar{e}$ along the directions $\bar{u}, \bar{m}, \bar{r}$ let $|\bar{u}| = |\bar{m}| = |\bar{r}| = K$
Let $\bar{e} = a \bar{u} + b \bar{m} + c \bar{r}$ $\bar{u} \times \{(\bar{e} - \bar{m}) \times \bar{u}\} = (\bar{u} . \bar{u}) (\bar{e} - \bar{m}) - (\bar{u} . (\bar{e} - \bar{m})) \bar{u}$
$\begin{array}{l} = K^{2} (\bar{e} - \bar{m}) - (\bar{u} . \bar{e}) \bar{u} (∵ \bar{u} . \bar{m} = \bar{m} . \bar{r} = \bar{m} . \bar{u} = 0) \\ = K^{2} (\bar{e} - \bar{m}) - (\bar{u} . \bar{e}) \bar{u} \end{array}$
Similar $\bar{m} \times \{(\bar{e} - \bar{r}) \times \bar{m}\} = (\bar{m} . \bar{m}) (\bar{e} - \bar{r}) - \bar{m} . (\bar{e} - \bar{r}) \bar{m}$
$= K^{2} (\bar{e} - \bar{r}) - (\bar{m} . \bar{e}) \bar{m}$ $\bar{r} \times \{(\bar{e} - \bar{u}) \times \bar{r}\} = (\bar{r} . \bar{r}) (\bar{e} - \bar{u}) - (\bar{r} . (\bar{e} - \bar{u})) . \bar{r}$
$= K^{2} (\bar{e} - \bar{u}) - (\bar{r} . \bar{e}) r$

Substitute above, in the given question $\bar{e} = \frac{1}{2} (\bar{u} + \bar{m} + \bar{r})$