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Let overset(to)(p) , overset(to)(q) , ov...

Let `overset(to)(p) , overset(to)(q) , overset(to)(r )` be three mutually perpendicular vectors of the same magnitude. If a vectors `overset(to)(X)` satisfies the equation
`overset(to)(p) xx [(overset(to)(x) -overset(to)(q)) xx overset(to)(p)] + overset(to)(q) xx [(overset(to)(x)-overset(to)(r ))xx overset(to)(q)]`
`+ overset(to)(r ) xx [(overset(to)(x) - overset(to)(p)) xx overset(to)(r )]=overset(to)(0) " then " overset(to)(x) ` is given by

A

`(1)/(2) (vec (p) +vec (q)-2vec (r ))`

B

`(1)/(2) (vec (p) +vec (q)+vec (r ))`

C

`(1)/(3) (vec (p) +vec (q)+vec (r ))`

D

`(1)/(2) (2vec (p) +vec (q)-vec (r ))`

Text Solution

Verified by Experts

The correct Answer is:
B
Since `vec(p) ,vec(q) ,vec(r )` are mutually perpendicular vectors of same magnitude so let us consider
`underset(vec(p) ". " vec(q)=vec(q) "." vec(r ) = vec(r ) ". " vec(p)=0)(|vec(p)|=|vec(q)|=|vec(r)|=lambda" and ")}`
Given `vec(p) xx {(vec(x)-vec(q)) xx vec(p)}+ vec(q) xx {(vec(x) -vec(r)) xx vec(q)}`
`+vec(r ) xx {(vec(x) - vec(p)) xx vec(r )}=vec(0)`
`rArr (vec(p) ". " vec(p)) xx (vec(X) - vec(q)) -{vec(p).(vec(X)-vec(q))}vec(p)+(vec(q) ". " vec(q)) (vec(X) -vec(r))`
`-{vec(q) ". " (vec(x) -vec(r )) } vec(q) + (vec(r).vec(r))(vec(x)-vec(p)) -{vec(r ).(vec(x) -vec(p))} vec(r )=0`
`rArr vec(X) {(vec(p)"." vec(p)) + (vec(q) "." vec(q))+ (vec(r) ". " vec(r))} - (vec(p) ". " vec(p))vec(q)`
`-(vec(q) ". " vec(q)) r- (vec(r ) ". " vec(r)) vec(p) = (vec(x)"." vec(q)) vec(p) + (vec(x ) ". " vec(q)) vec(q) + (vec(x) ". " vec(r)) vec(r)`
`rArr 3 vec(X) |lambda|^(2) -(vec(p) +vec(q))|vec(r))|lambda|^(2) =(vec(X)"." vec(p))vec(p)`
`+(vec(x ) ". " vec(q)) vec(q) + (vec(x) ". " vec(r)) vec(r)`
Taking dot of Eq. (ii) with `vec(p)` we get
`3(vec(X)"." vec(p)) |lambda|^(2) -|lambda|^(4) =(vec(x ) ". " vec(p)) |lambda|^(2) rArr vec(x) ". " vec(p) = (1)/(2) |lambda|^(2)`
Similarly taking dot of Eq . (ii) with `vec(q) " and " vec(r )` respectively we get
` vec(x) ". " vec(q) = (|lambda|^(2))/(2) = vec(x) ". " vec(r )`
`:. Eq ` (ii) becomes
`3vec(x ) |lambda|^(2) -(vec(p) +vec(q) + vec(r )) |lambda|^(2) = (|lambda|^(2))/(2) (vec(p) +vec(q) + vec(r ))`
`rArr 3vec(x) = (1)/(2) (vec(p) +vec(q) +vec(r )) + (vec(p) +vec(a) +vec(r ))`
`rArr vec(X) = (1)/(2) (vec(p)+vec(q) +vec(r ))`
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