If `vec(A)=2hati+hatj+hatk " and " vecB=10hati+5hatj+5hatk`, if the magnitude of component of `(vec(B)-vec(A))` along `vec(A)` is `4sqrt(x)`. Then x w

Correct Answer - (i) `sin^(-1)((2sqrt(2))/(3))`, (ii) `sin^(-1)((3sqrt(3))/(sqrt(29)))`, (iii) `sin^(-1)((-4)/(sqrt(406)))` , (iv) `sin^(-1)((23)/(11sqrt(11)))`

Here, `veca_(1) = hati+2hatj+hatk, vecb_(1)= hati-hatj+hatk` `veca_(2)=2hati-hatj-hatk, vecb_(2)=2hati+hatj+2hatk` `:. veca_(2)-a_(1)=hati-3hatj-2hatk` `vecb_(1) xx vecb_(2) = |{:(hati,hatj,hatk),(1,-1,1),(2,1,2):}| = - 3 hati+3hatk` Shortest distance between two lines `= |{:((veca_(2)-veca_(1)).(vecb_(1)xxb_(2)))/(|vecb_(1)xxvecb_(2)|):}|` `= |{:((hati-3hatj-2hatk).(-3hati+3hatk))/(|-3hati+3hatk|):}|` `= |{:(-3-6)/(sqrt(9+9)):}|=(9)/(3sqrt(12))...

Correct Answer - a Equation of the plane containing `L_(1),A(x-2)+B(y-1)+C(z+1)=0` where `A+2C=0,A+B-C=0` `impliesA=-2C-,B=3C,C=C` `implies"Plane is"-2(x-2)+3(y-1)+z+1=0` or `2x-3y-z-2=0` Hence, `p=|(-2)/(sqrt14)|=sqrt((2)/(7))`

Correct Answer - b The required line passes through the point `hati+3hatj+2hatk` and is perpendicular to the lines `vecr= (hati+2hatj-hatk)+ lamda(2hati+hatj+hatk) and vecr= (2hati+ 6hatj+hatk)+mu (hati+ 2hatj+3hatk)`, therefore, it is parallel...

Correct Answer - b Given lines are parallel as both are directed along the same vector `(hati+hatj-hatk)` , so they do not intersect. Also Statement 2 is correct by definitioin of...

`|{:(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`

`|{:(a_(1) , a_(2),a_(3)),(b_(1),b_(2),b_(3)),(c_(1),c_(2),c_(3)):}|^(2)= [veca vecb vecc]^(2)` ` = ((veca xx vecb) .vecc)^(2)` ` ( ab sin theta vecc. vecc)^(2)` . `(a^(2)b^(2))/4` ` 1/4 (a_(1)^(2) + a_(2)^(2)+a_(3)^(2))(b_(1)^(2) =b_(2)^(2) + b_(3)^(2)) `

Correct Answer - 7 `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` L.H.S = ` [3veca +vecb 3vecb+vecc 3vecc +veca]` `[3veca 3vecb 3vecc] + [vecb vecc veca]` `3^(3)[veca vecb vecc] +...

Correct Answer - c Any vector coplanar, to `veca and vecb ` can be written as `vecr=muveca + lamdavecb` `vecr = (mu + 2lamda) hati + ( -mu+lamda) hatj + (...

Correct Answer - a A vector in the plane of `veca and vecb` is `vecu=muveca +lambdavecb= (mu +lamda)hati+ (2mu - lamda) hatj + ( mu +lambda)hatk` projection of `vecu` on `...