If `veca, vecb and vecc` are three non-coplanar non-zero vectors, then prove that `(veca.veca) vecb xx vecc + (veca.vecb) vecc xx veca + (veca.vecc)veca xx vecb = [vecb vecc veca] veca`
1 Answers
As `veca ,vecb and vecc` are non- coplanar, `vecb xx veca,veccxxveca and vecaxxvecb` are also non-coplanar, So, any vector con be expressend as a linear combination of these vectors.
`veca=lambdavecbxxvecc+mu vecc xx veca+ v vecaxxvecb`
`veca.veca=lambda[vecbveccveca],veca.vecb=mu [vecc vecavecb], veca.vecc=v[vecavecbvecc]`
`veca=((veca.veca)vecbxxvecc)/([vecb veccveca])+((veca.vecb)veccxxveca)/([veccvecavecb])+((veca.vecc)vecaxxvecb)/([vecavecbvecc])`
0
like
0 dislike
Talk Doctor Online in Bissoy App
The magnitudes of vectors `vecA.vecB` and `vecC` are respectively 12,5 and 13 unira and `vecA+vecB=vecC`, then the angle between `vecA` and `vecB` is
Correct Answer - C `vecA+vecB=vecC` as `vecA^(2)+B^(2)=C^(2)` thus angle between A and B is `90^(@)`
2 Answers 1 views
If `veca, vecb and vecc` are vectors such that `|veca|=3,|vecb|=4 and |vec|=5 and (veca+vecb)` is perpendicular to vecc,(vecb+vecc) is perpendicular t
Given , `(veca+vecb)=0Rightarrowveca.vecc+vecb.vecc=0` `(vecb+vecc).veca=0Rightarrowveca.vecb+vecc.veca=0` `(vecc+veca).vecb=0Rightarrowvecb.vecc+veca.vecb=0` `2(veca.vecb+vecb.vecc+vecc.veca)=0``Now, |veca+vecb+vecc|^(2)+|vecb|^(2)+|vecc|^(2)+2(veca.vecb+vecb.vecc+vecc.veca)=50` `|veca+vecb+vecc|=5sqrt2`
2 Answers 1 views
If `veca, vecb and vecc` are three non-coplanar vectors, then find the value of `(veca.(vecbxxvecc))/(vecb.(veccxxveca))+(vecb.(veccxxveca))/(vecc.(ve
Since ` (veca vecb vecc] ne 0`, we have `(veca.(vecbxxvecc))/(vecb.(veccxxveca))+(vecb.(veccxxveca))/(vecc.(vecaxxvecb))+(vecc.(vecbxxveca))/(veca.(vecbxxvecc))=([veca vecb vecc])/([vecb vecc veca])+([vecbveccveca])/([vecc veca vecb])+([vecc vecb veca])/([veca vecb vecc])` `= ([vecavecb vecc])/([veca vecbvecc])+([vecavecbvecc])/([vecavecbvecc])-([vecavecbvecc])/([veca vecb vecc])` = 1+ 1 -1=1
2 Answers 1 views
Let `veca, vecb, vecc` be three unit vectors and `veca.vecb=veca.vecc=0` . If the angle between `vecb and vecc` is `pi/3` then find the value of `|[ve
`veca.vecb=veca. Vecc=0` is perpendicular to vectors `vecb and vecc`. Thus `veca=lambda(vecbxxvecc)` `|veca|=|lambda(vecbxxvecc)=|lambdasqrt3/2|=1` ` |[veca vecbvecc]|=|veca.(vecbxxvecc)|` `=|lambda||(vecbxxvecc)|^(2)` `=2/sqrt3|vecb|^(2)|vecc|^(2)sin^(2)(pi/3)=2/sqrt3xx(sqrt3/2)^(2)=sqrt3/2`
2 Answers 1 views
Show that : `[vecl vecm vecn] [veca vecb vecc]=|(vecl.veca, vecl.vecb, vecl.vecc),(vecm.veca, vecm.vecb, vecm.vecc),(vecn.veca, vecn.vecb, vecn.vecc)|
Let `vecl=l_(1)hati+l_(2)hatj+l_(3)hatk,vecm=m_(1)hati+m_(2)hatj+m_(3)hatk andvecn=n_(1)hati+n_(2)hatj+n_(3)hatk` `veca=a_(1)hati+a_(2)hatj=a_(3)hatk,vecb=b_(1)hati+b_(2)hatj+b_(3)hatk nandvecc=c_(1)hati+c_(2)hatj+c_(3)hatk` `vecl.veca=l_(1)a_(1)+l_(2)a_(2)+l_(3)a_(3)=suml_(1)a_(1)` similarly `vecl.vecb=suml_(1)b_(1).etc.` `[veclvecmvecn][vecavecbvecc]=|{:(l_(1),l_(2),l_(3)),(m_(1),m_(2),m_(3)),(n_(1),n_(2),n_(3)):}||{:(a_(1),a_(2),a_(3)),(b_(1),b_(2),b_(3)),(c_(1),c_(2),c_(3)):}|` `=|{:(suml_(1)a_(1),suml_(1)b_(1),suml_(1)c_(1)),(summ_(1)a_(1),summ_(2)b_(1),summ_(1)c_(1)),(sumn_(1)a_(1),sumn_(1)b_(1),sumn_(1)c_(1)):}|` `=|{:(vecl.veca,vecl.vecb,veca.vecc),(vecm.veca,vecm.vecb,vecm.vecc),(vecn.veca,vecn.vecb,vecn.vecc):}|`
2 Answers 1 views
If `veca=hati+hatj+hatk,hatb=hati-hatj+hatk,vecc=hati+2hatj-hatk`, then find the value of `|{:(veca.veca,veca.vecb,veca.vecc),(vecb.veca,vecb.vecb,vec
`|{:(veca.veca,veca.vecb,veca.vecc),(vecb.veca,vecb.vecb,vecb.vecc),(vecc.veca,vecc.vecb,vecc.vecc):}|=[veca vecbvecc][vecavecbvecc]=[vecavecbvecc]^(2)` `|{:(veca.veca,veca.vecb,veca.vecc),(vecb.veca,vecb.vecb,vecb.vecc),(vecc.veca,vecc.vecb,vecc.vecc):}|= 4^(2)=16`
2 Answers 1 views
If `veca,vecb and vecc` are non coplnar and non zero vectors and `vecr` is any vector in space then`[vecc vecr vecb]veca+pveca vecr vecc] vecb+[vecb v
`Let" " vecr=x_(1)veca+x_(2)vecb+x_(3)veccRightarrow vecr.(vecbxxvecc)=x_(1)veca.(vecbxxvecc)or x_(1)=([vecr vecb vecc])/([veca vecb vecc])`. `Also, " " vecr.(veccxxveca)=x_(2)vecb.(veccxxveca)orx_(2)=([vecrveccveca])/([vecavecbvecc])` `and vecr.(vecaxxvecb)=x_(3)vecc. (vecaxxvecb)or x_(3)=([vecr vecavecb])/([veca vecbvecc])` `Rightarrow vecr=([vecrvecb vecc])/([vecavecbvecc])veca+([vecrveccveca])/([vecavecbvecc])vecb+([vecrvecavecb])/([vecavecb vecc])vecc` `or [vecbveccvecr]veca+[veccvecavecr]vecb+[vecavecbvecr]vecr=[vecavecbvecc]vecr`
2 Answers 1 views
Given `|veca|=|vecb|=1 and |veca + vecb|= sqrt3 "if" vecc` is a vector such that `vecc -veca - 2vecb = 3(veca xx vecb) ` then find the value of `vecc
We, have, `|veca + vecb|=sqrt3` ` or |veca + vecb|^(2) =3` `or |veca|^(2) + |vecb|^(2) + 2(veca .vecb) =3` `or veca. Vecb = 1//2` (i) now ` vecc - veca...
2 Answers 2 views
Let `veca,vecb and vecc` be the non zero vectors such that `(vecaxxvecb)xxvecc=1/3 |vecb||vecc|veca.` if theta is the acute angle between the vectors
we have `(veca xx vecb ) xx vecc = 1/3 |vecb||vecc|veca` `or (veca .vecc) vecb - (vecb .vecc)veca = 1/3 |vecb||vecc|veca` `or (veca. Vecc) vecb- {(vecb.vecc) + 1/3 |vecb|vecc|} veca...
2 Answers 1 views
`vecb and vecc` are non- collinear if `veca xx (vecb xx vecc) + (veca .vecb) vecb = ( 4-2x- sin y) vecb + ( x^(2) -1) vecc nad (vec. Vecc) veca =veca
Correct Answer - a,c ` veca xx (vecb xx vecc) + (veca .vecb)vecb` `= ( 4 - 2 x - sin y) vecb + ( x^(2) -1) vecc` ` or (veca .vecc0...
2 Answers 1 views