The centre of the circle given by `vecr.(hati+2hatj+2hatk)=15and|vecr-(hatj+2hatk)=4` is

The distance between the planes given by vecr.(hati+2hatj-2hatk)+5=0 and vecr.(hati+2hatj-2hatk)-8=0 is

Equation of a plane passing through the intersection of the planes vecr.(3hati-hatj+hatk)=1 and vecr.(hati+4hatj-2hatk)=2 and passing through the point (hati+2hatj-hatk) is :

The vector parallel to the line of intersection of the planes vecr.(3hati-hatj+hatk) = 1 and vecr.(hati+4hatj-2hatk)=2 is :

The plane through the point (-1,-1,-1) nd containing the line of intersection of the planes vecr.(hati+3hatj-hatk)=0 ,vecr.(hatj+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

Find the line of intersection of the planes vecr.(3hati-hatj+hatk)=1 and vecr.(hati+4hatj-2hatk)=2

The angle between the planes vecr. (2 hati - 3 hatj + hatk) =1 and vecr. (hati - hatj) =4 is

Show that the line of intersection of the planes vecr*(hati+2hatj+3hatk)=0 and vecr*(3hati+2hatj+hatk)=0 is equally inclined to hati and hatk . Also find the angleit makes with hatj .

Prove that the equaton of a plane through point (2,-4,5) and the line o lintersection of the planes vecr.(2hati+3hatj-4hatk) = 1 and vecr.(3hati+hatj-2hatk) = 2 is vecr.(2hati+8hatj+7hatk) = 7 .

The angle between the planes vecr.(2hati-hatj+hatk)=6 and vecr.(hati+hatj+2hatk)=5 is

Find the equation of the plane through the point hati+4hatj-2hatk and perpendicular to the line of intersection of the planes vecr.(hati+hatj+hatk)=10 and vecr.(2hati-hatj+3hatk)=18.