Given two orthogonal vectors `vecA and vecB` each of length unity. Let `vecP` be the vector satisfying the equation `vecP xxvecB=vecA-vecP`. then
`vecP` is equal to

Options

A. `vecA/2 + (vecAxxvecB)/2`
B. `vecA/2 + (vecBxxvecA)/2`
C. `(vecAxxvecB)/2 -vecA/2`
D. `vecA xxvecB`

Answer

b

Explanation

`vecPxxvecB=vecA-vecP and |vecA|=|vecB|=1 and vecA.vecB=0`is given
Now `vecP xx vecB = vecA-vecP`
`(vecPxxvecB) xxvecB= (vecA-vecP)xxvecB`
(taking cross product with `vecB` on both sides)
`or (vecP.vecB)vecB- (vecB.vec B)vecP=vecAxxvec B -vecP xx vecB`
`or (vecP.vecB) vecB-vecP=vecA xx vecB-vecA+vecP`
`or 2vecP = vecA-vecA-vecAxxvec B-(vecP.vecB)vecB`
`or vecP= (vecA-vecAxxvecB=-(vecP.vecB)vecB)/2`
taking dot product with `vecB` on both sides of (i) get
`vecP.vecB=vecA.vecB-vecP.vecB`ltbr gt `Rightarrow vecP.vecB=0`
`Rightarrow vecP=( vecA + vecBxxvecA)/2`
Now
`(vecP xx vecB) xx vecB= (vecP.vecB) vecB-(vecB.vecB) vecP=-vecP`
`vecP,vecA,vecPxxvecB(=vecA-vecP)` are depenment
Also `vecP.vecB=0`
` and |vecP|^(2)=|(vecA-vecAxxvecB)/2|^(2)`
`(|vecA|^(2)+|vecAxxvecB|^(2))/4`
`(1+1)/4=1/2or |vecP|=1/sqrt2`