समतलो `vecr*(hati+2hatj+hatk)` तथा `hatr*(2hati+hatk-hatk)+5=0` के बिच का कोण ज्ञात कीजिए !

If the planes vecr.(hati + hatj + hatk) = q _(1) , hatr. (hati + 2 a hatj + hatk) = q _(2) and vecr.(ahati + a ^(2) hatj + hatk) = q _(3) intersect in a line, then the value of a is

Find the angle between the line vecr = (hati +2hatj -hatk ) +lambda (hati - hatj +hatk) and vecr cdot (2hati - hatj +hatk) = 4.

Find the shortest distance between the lines vecr = (hati + 2hatj+hatk) + lambda(hati-hatj+hatk) and vecr = 2hati - hatj-hatk + mu(2hati+hatj+2hatk)

The angle between the line vecr=( hati+2 hatj- hatk)+ lambda( hati- hatj+ hatk) and the plane vecr. (2 hati- hatj+ hatk)=4 is __

The shortest distance between the lines vecr = (2hati - hatj) + lambda(2hati + hatj - 3hatk) vecr = (hati - hatj + 2hatk) + lambda(2hati + hatj - 5hatk)

Find the acute angle between the following planes: i) vecr.(hati+hatj-2hatk)=5 and vecr.(2hati+2hatj-hatk)=9 ii) vecr.(hati+2hatj-hatk)=6 and vecr.(2hati-hatj-hatk)+3=0

The plane through the point (-1,-1,-1) nd contasining the line of intersection of the planes vecr.(hati+3hatj-hatk)=0 ,vecr.(hati+2hatk)=0 is (A) vecr.(hati+2hatj-3hatk)=0 (B) vecr.(hati+4hatj+hatk)=0 (C) vecr.(hati+5hatj-5hatk)=0 (D) vecr.(hati+hatj+3hatk)=0