मूल बिंदु से समतल `overline(r).(2hat(i) + hat(j) + 2hat(k)) = 6` की दूरी ज्ञात कीजिए।

If angle bisector of overline(a) = 2hat(i) + 3 hat(j) + 4 hat(k) and overline(b) = 4hat(i) - 2 hat(j) + 3 hat(k) is overline(c ) = alpha hat(i) + 2 hat(j) + beta hat(k) then

If overline(a)*hat(i)=4, then (overline(a)timeshat(j))*(2hat(j)-3hat(k))=

A plane is at a distance of 5 units from the origin and perpendicular to the vector 2hat(i) + hat(j) + 2hat(k) . The equation of the plane is a) vec(r ).(2hat(i) + hat(j) -2hat(k))=15 b) vec(r ).(2hat(i)+ hat(j)-hat(k))=15 c) vec(r ).(2hat(i) + hat(j)+ 2hat(k))=15 d) vec(r ).(hat(i) + hat(j) + 2hat(k))=15

Find the shortest distance between the lines : vec(r) = (4hat(i) - hat(j)) + lambda(hat(i) + 2hat(j) - 3hat(k)) and vec(r) = (hat(i) - hat(j) + 2hat(k)) + mu (2hat(i) + 4hat(j) - 5hat(k))