1 .Find the equation of the plane passing through the line of the plane and 2x+5y+3z=9 and parallel to the line 6x=2y =3z. Sol: Equation of required plane:2x+2y-3z-7+λ(2x+5y+3z-9)=0 (2+2λ )x+(2+5λ)y+(-3+3λ)z+(-7-9λ)=0 (1) Since the plane is parallel to the line x/1=y/3=z/2 (2+2λ)1+(2+5λ)2+(-3+3λ)2=0 Λ=-2/23. Substituting the value of λ in eq. 1 Ans: 42x+36y-75z-143=0 is the required equation. 2. Find the equation of plane passing through the line of intersection of the planes x+2y+3z =4 and 2x+y-z+5 =0 and perpendicular to the plane 5x+3y-6z+8 =0. Sol: The required plane is (x+2y + 3z ) + k (2x + y – z +5 )= 0 Or (1+2k)x +(2+k)y +(3-k)z-4+5k=0 : Since it is perpendicular to 5x +3y-6z+8=0, 5(1+2k) +3 (2+k) -6 (3-k)=0, i.e k= 7/19, The equation of the plane is: 33x+45y +50z = 41 . Find the shortest distance between the lines : . 3. Sol: Here So, = 3 And
1 .Find the equation of the plane passing through the line of the plane and 2x+5y+3z=9 and parallel to the line 6x=2y =3z. Sol: Equation of required plane:2x+2y-3z-7+λ(2x+5y+3z-9)=0 (2+2λ )x+(2+5λ)y+(-3+3λ)z+(-7-9λ)=0 (1) Since the plane is parallel to the line x/1=y/3=z/2 (2+2λ)1+(2+5λ)2+(-3+3λ)2=0 Λ=-2/23. Substituting the value of λ in eq. 1 Ans: 42x+36y-75z-143=0 is the required equation. 2. Find the equation of plane passing through the line of intersection of the planes x+2y+3z =4 and 2x+y-z+5 =0 and perpendicular to the plane 5x+3y-6z+8 =0. Sol: The required plane is (x+2y + 3z ) + k (2x + y – z +5 )= 0 Or (1+2k)x +(2+k)y +(3-k)z-4+5k=0 : Since it is perpendicular to 5x +3y-6z+8=0, 5(1+2k) +3 (2+k) -6 (3-k)=0, i.e k= 7/19, The equation of the plane is: 33x+45y +50z = 41 . 3. Find the shortest distance between the lines : . Sol: Here So, = 3 And
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