Let ` vec a , vec b ,a n d vec c` be non-coplanar unit vectors, equally inclined to one another at an angle`theta` . If ` vec axx vec b+ vec bxx vec c=p vec a+q vec b+r vec c ,` find scalars `p ,qa n dr` in terms of `thetadot`

Answer

`p=r= (1)/(sqrt(1+2 cos 0) , q= (-2cos 0)/(sqrt(1+2 cos 0)`

Explanation

Since `vec(a) ,vec(b), vec(c )` are non- coplanar vectors .
`rArr [vec(a) vec(b) vec( c)] ne 0`
Also `vec(a) xx vec(b) + vec(b) xx vec( c) = p vec(a) + q vec(b) + r vec(c )`
Taking dot product with `vec(a) , vec(b) " and " vec( c)` respectively both sides we get
`p+q cos theta + r cos theta = [vec(a) vec(b) vec( c)] " "......(i)`
`p cos theta + q r cos theta =0" " .....(ii)`
and `p cos 0 + q cos theta + r = [ vec(a) vec( b) vec(c )] " " .....(ii)`
On adding above equations
`p+q + r= (2 [vec(a) vec(b) vec(c )])/(2 cos theta +1) " ".....(iii)`
On multiplying Eq. (iv) by cos `theta` and subtracting Eq. (i) we get
` p (cos theta -1) = (2[vec(a) vec(b)vec(c )])/(2 cos theta +1) -[vec(a) vec( b) vec( c)]`
`rArr p = ([vec(a) vec(b ) vec( c)])/((1-cos theta )(1-cos theta))`
Similarly `q= (-2[vec(a) vec(b) vec( c)] cos theta)/((1+2 cos theta )(1 - cos theta))`
and `r= ([vec(a)vec(b) vec(c )])/((1+2 cos theta) (1-cos theta))`
Now `[vec(a) vec(b )vec( c)]^(2)= |{:(vec(a)"."vec(a),,vec(a)"."vec(b),,vec(a)"."vec(b)),(vec(b)"."vec(a),,vec(b)"."vec(b),,vec(b)"."vec(c)),(vec(c)"."vec(a),,vec(c) "."vec(b),,vec(c)"."vec(c)):}|= |{:(1,,cos theta,,cos theta),(cos theta,,1,,cos theta),(cos theta,,cos theta,,1):}|`
Applying `R_(1) to R_(1) + R_(2) +R_(2)`
`=(1+2 cos theta ) |{:(1,,1,,1),(cos theta,,1,,cos theta),(cos theta ,,cos theta ,,1):}|` ltbr. Applying `C_(2) to C_(2) -C_(1) ,C_(3) to C_(3) -C_(1)`
`=(1+ 2 cos theta ) .|{:(1,,0,,0),(cos theta ,,1-cos theta ,,0),(cos theta ,,0 ,,1-cos theta):}|`
` =(1+2 cos theta) . (1-cos theta)^(2)`
`rArr [vec(a) vec(b) vec(c )] = (sqrt(1+2 cos theta ).(1-cos theta )`
`:. P = (1)/(sqrt(1+2 cos theta) ) , q = (-2 cos theta)/(sqrt(1+2cos theta)) " and " r= (1)/(sqrt(1 +2cos theta))`