CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Find the equation of the plane passing through the point (1,3,2) and perpendicular to each of the planes x+2y+3z=5 and 3x+3y+z=0.

Open in App
Solution

The equation of the plane passing through (1,2,3) will be given by a(x+1)+b(y2)+c(z3)=0, where a,b and c are the direction ratios of the normal to the plane.

Also, this plane is perpendicular to the planes x+2y+3z=5 and 3x+3y+z=0. According to the perpendicularity condition of planes, the dot product of the direction ratios of the normals to the two planes should be 0.

a+2b+3c=0 and 3a+3b+c=0

c=a+2b3

Substituting this in the second equation,

3a+3ba+2b3=0

9a+9ba2b=0

8a+7b=0

b=8a7

c =a+2b3

=a16a73

=9a7×3

=3a7

Substituting these values of b and c in the equation of the plane,

a(x+1)8a7(y2)+3a7(z3)=0

7(x+1)8(y2)+3(z3)=0

7x8y+3z+7+169=0

7x8y+3z+14=0

Hence, the equation of the required plane is 7x8y+3z+14=0

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Tango With Straight Lines !!
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon