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The vector equation of a plane is `vecr.(3hati+2hatj-6hatk) = 56`. Convert it into normal form. Also find the length of perpendicular from origin and direction cosines of normal to the plane. |
Answer» `vecr.(3hati+2hatj-6hatk) = 56` Here `vecn=3hati+2hatj-6hatk` `rArr |vecn |=sqrt(9+4+36) = 7` `:. vecr.((3hati+2hatj-6hatk))/(7) = (56)/(7)` `rArr vecr.(3/7hati+2/7hatj-6/7hatk) = 8` Which is the normal form of the plane. Direction cosines of the perpendicular drawn from ` =3/7,2/7,(-6)/(7)` origin and length of perpendicular from origin `= 8` units. |
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