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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 toA. `vecA/2 + (vecAxxvecB)/2`B. `vecA/2 + (vecBxxvecA)/2`C. `(vecAxxvecB)/2 -vecA/2`D. `vecA xxvecB` |
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Answer» Correct Answer - b `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` |
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