Home
Class 11
MATHS
The d.c.'s of the normal to the plane 2x...

The d.c.'s of the normal to the plane 2x + 3y – 6z + 5 = 0 are

Text Solution

Verified by Experts

The correct Answer is:
3
Promotional Banner

Topper's Solved these Questions

  • THE PLANE

    VIKRAM PUBLICATION ( ANDHRA PUBLICATION)|Exercise TEXTUAL EXERCISES (EXERCISE 7(a) (II))|8 Videos
  • THE PLANE

    VIKRAM PUBLICATION ( ANDHRA PUBLICATION)|Exercise TEXTUAL EXERCISES (EXERCISE 7(a) (III))|6 Videos
  • THE PLANE

    VIKRAM PUBLICATION ( ANDHRA PUBLICATION)|Exercise IMPORTANT POINTS |4 Videos
  • TELANGANA MARCH 2015

    VIKRAM PUBLICATION ( ANDHRA PUBLICATION)|Exercise Section - C (Long answer type questions)|6 Videos
  • THREE DIMENSIONAL CO-ORDINATES

    VIKRAM PUBLICATION ( ANDHRA PUBLICATION)|Exercise TEXTUAL EXERCISES (EXERCISE -5(b))|10 Videos

Similar Questions

Explore conceptually related problems

Find the d.c's of the normal to the plane 2x+4y+5z=1 .

Find the d.c's of the normal to the plane 2x+4y+5z=1 .

Find d.c's of the normal to plane 3x - 6y +2z -4 = 0

The intercepts of the plane 2x – 3y + 5z – 30 = 0 are

Find the d.r.s and d.c.s of normal to the plane 3x-2y+6z-7=0 .

Find the d.r.s and d.c.s of normal to the plane 3x-2y+6z-7=0 .

The perpendicular distance from the point (-2,3, 1) to the plane 2x – 3y – 6z + 5 = 0 is

The D.R's of a normal to the planes are 1, 2, 3 and distance of plane from (0, 0, 0) is 5, then equation of plane is :

The equations of the bisectors of the angles between the planes 3x – 6y + 2z + 5 = 0, 4x – 12y + 3z - 3 = 0 is

The d.r's of the normal to the plane passing through the point (2,1,3) and the line of intersection of the planes x+2y+z=3 and 2x-y-z=5 is