answer questions 2y-2z + 4 = 0 .. - by + 42 = 8 __. Question 28. planes: i + 2 + 3k) - 4 = 0 and | r • (2 i + j-k) + 5 = 0 containing the intersection line And perpendicular plane of plane r • (5 i + 3 j - 6k) + 8 = 0 Find the equation. [NCERT EXERCISE) Solution: Let the planes • i + 21-27 r (31 Known solution r • (3 i → AMA

A unit vector perpendicular to the plane of a = 2i - 6j -3k, b = 4i + 3j - k is

The distance between the planes r. (i+ 2j - 2k ) + 5 =0 and r. (2i + 4j - 4k ) - 16 =0 is

Find the direction cosines of perpendicular from the origin to the plane vec r*(2hat i-3hat j-6hat k)+5=0

The angle between the planes r. (2i - j + 2k) = 3 and r. (3i - 6j + 2k) = 4

Equation of the line passing through i+j -3k and perpendicular to the plane 2x - 4y + 3z + 5 =0 is -

(i) Find the vector equation of the plane through the intersection of the planes : vec(r) . (hati + hatj + hatk) = 6, vec(r) . (2 hati + 3 hatj + 4 hatk) = -5 and the point (1,1,1). (ii) Find the equation of the plane which contains the line of intersection of the planes : vec(r) . (hat(i) + 2 hat(j) + 3 hat(k) ) - 4 = 0. vec(r). (2 hat(i) + hatj - hat(k) ) + 5 = 0 and which is perpendicular to the plane : vec(r) . (5 hati + 3 hatj - 6 hatk ) + 8 = 0 . (iii) Find the equation the plane passing through the intersection of the planes x + y + z = 6 and 2x + 3y + 4z + 5 = 0 and the point (1,1,1) .

The line of intersection of the planes vec r . (3i-j+k) =1 and vec r . (i+4j-2k) = 2 is parallel to the vector