Position vector ` hat k`is rotated about the originby angle `135^0`in such a way that theplane made by it bisects the angel between ` hat ia n d hatjdot`Then its new position is`+-( hat i)/(sqrt(2))+-( hat j)/(sqrt(2))`b. `+-( hat i)/2+-( hat j)/2-( hat k)/(sqrt(2))`c. `( hat i)/(sqrt(2))-( hat k)/(sqrt(2))`d. none of theseA. `+-hati/sqrt2+-hatj/sqrt2`B. `+-hati/2+-hatj/2-hatk/sqrt2`C. `hati/sqrt2-hatk/sqrt2`D. none of these

Position vector ` hat k`is rotated about the originby angle `135^0`in

1.	Position vector ` hat k`is rotated about the originby angle `135^0`in such a way that theplane made by it bisects the angel between ` hat ia n d hatjdot`Then its new position is`+-( hat i)/(sqrt(2))+-( hat j)/(sqrt(2))`b. `+-( hat i)/2+-( hat j)/2-( hat k)/(sqrt(2))`c. `( hat i)/(sqrt(2))-( hat k)/(sqrt(2))`d. none of theseA. `+-hati/sqrt2+-hatj/sqrt2`B. `+-hati/2+-hatj/2-hatk/sqrt2`C. `hati/sqrt2-hatk/sqrt2`D. none of these
Answer» Correct Answer - d Let `vecr` be the new position . Then `vecr= lambdahatk + mu ( hati + hatj) ` ` Also, vecr.hatk = - 1/sqrt2 Rightarrow lambda= - 1/sqrt2` ` Also lambda^(2)+ 2mu^(2) = 1 Rightarrow 2mu^92) = 1/2 or mu = +- 1/2` ` vecr = +- 1/2 (hati + hatj) =- hatk/sqrt2`

Discussion

No Comment Found

Related InterviewSolutions

If `veca=a_(1)hati+a_(2)hatj+a_(3)hatk, vecb= b_(1)hati+b_(2)hatj + b_(3)hatk, vecc=c_(1)hati+c_(2)hatj+c_(3)hatk and [3veca+vecb 3vecb+vecc 3vecc + veca] =lambda|{:(veca.hati,veca.hatj,veca.hatk),(vecb.hati,veca.hatj,hatb.hatk),(vecc.hati,vecc.hatj,vecc.hatk):}| " then find the value of " lambda/4`
`veca, vecb and vecc` are unit vecrtors such that `|veca + vecb+ 3vecc|=4` Angle between `veca and vecb is theta_(1)` , between `vecb and vecc is theta_(2)` and between `veca and vecb` varies `[pi//6, 2pi//3]` . Then the maximum value of `cos theta_(1)+3cos theta_(2)` isA. 3B. 4C. `2sqrt2`D. 6
The volume of a tetrahedron fomed by the coterminus edges `veca , vecb and vecc is 3` . Then the volume of the parallelepiped formed by the coterminus edges `veca +vecb, vecb+vecc and vecc + veca` isA. 6B. 18C. 36D. 9
Let `vecb=4hati+3hatj and vecc` be two vectors perpendicular to each other in the xy-plane. Find all vetors in te same plane having projection 1 and 2 along `vecb and vecc` respectively.
Let `veca and vecb` be unit vectors that are perpendicular to each other l. then `[veca+ (veca xx vecb) vecb + (veca xx vecb) veca xx vecb]` will always be equal toA. 1B. 0C. `-1`D. none of these
Vectors `vecx,vecy,vecz` each of magnitude `sqrt(2)` make angles of `60^0` with each other. If `vecxxx(vecyxx(veczxxvecx)=vecb nd vecxxxvecy=vecc, find vecx, vecy, vecz` in terms of `veca,vecb and vecc`.A. `1/2 [(veca + vecc)xxvecb- vecb -veca]`B. `1/2 [(veca - vecc)xxvecb+ vecb +veca]`C. `1/2 [(veca - vecb)xxvecc+ vecb +veca]`D. `1/2[(veca-vecc)xx veca + vecb -veca]`
Vectors `vecx,vecy,vecz` each of magnitude `sqrt(2)` make angles of `60^0` with each other. If `vecxxx(vecyxx(veczxxvecx)=vecb nd vecxxxvecy=vecc, find vecx, vecy, vecz` in terms of `veca,vecb and vecc`.A. `1/2[(veca-vecc) xx vecc-vecb+veca]`B. `1/2[(veca-vecb) xx vecc+vecb-veca]`C. `1/2[veccxx(veca-vecb) + vecb +veca]`D. none of these
Let `veca=hati+4hatj+2hatk,vecb=3hati-2hatj+7hatk and veca=2hati-2hatj+4hatk` . Find a vector `vecd` which perpendicular to both `veca and vecb and vecc.vecd=15`.
The value of a so thast the volume of parallelpiped formed by vectors `hati+ahatj+hatk, hatj+ahatk, ahati+hatk` becomes minimum is (A) `sqrt93)` (B) 2 (C) `1/sqrt(3)` (D) 3
If `vecr and vecs` are non-zero constant vectors and the scalar b is chosen such that `|vecr+bvecs|` is minimum, then the value of `|bvecs|^(2)+|vecr+bvecs|^(2)` is equal toA. `2|vecr|^(2)`B. `|vecr|^(2)//2`C. `3|vecr|^(2)`D. `|vecr|^(2)`

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment