Prove that the normals to the planes 4x + 11y + 2z + 3 = 0 and 3x - 2y + 5z = 8 are perpendicular to each other.
Given :
Equations of plane are :
4x + 11y + 2z + 3 = 0
3x – 2y + 5z = 8
To Prove : are perpendicular.
Formula :
1) Dot Product :
If are two vectors
then,
Note :
Direction ratios of the plane given by
ax + by + cz = d
are (a, b, c).
For plane
4x + 11y + 2z + 3 = 0
direction ratios of normal vector are (4, 11, 2)
therefore, equation of normal vector is
And for plane
3x – 2y + 5z = 8
direction ratios of the normal vector are (3, -2, 5)
therefore, the equation of normal vector is
Now,
= 12 – 22 + 10
= 0
Therefore, normals to the given planes are perpendicular.