Equation of a Plane : Point Normal Form
Trending Questions
Q. If direction ratios of the normal of the plane which contains the line x−23=y−42=z−11 is (a, 1, −26), then a=
Q. The equation of a plane passing through (1, 2, –3), (0, 0, 0) and perpendicular to the plane 3x – 5y + 2z = 11, is :
- 4x + y + 2z = 0
- x + y + z = 0
- 3x+y+53z=0
- 3x−y+z3=0
Q. The equation of the plane passing through the point (–2, –2, 2) and containing the line joining the points (1, 1, 1) and (1, –1, 2) is :
- x + 2y – 3z + 4 = 0
- 5x + 2y – 3z – 17 = 0
- x – 3y – 6z + 8 = 0
- 3z – 4y + 1 = 0
Q. P is a fixed point (a, a, a) on a line through the origin equally inclined to the axes, then any plane through P perpendicular to OP, makes intercepts on the axes, the sum of whose reciprocals is equal to
- a
- 32a
- 3a2
- 1a
Q. Equation of the plane perpendicular to the plane x – 2y + 5z + 1 = 0 which passes through the points (2, –3, 1) and (–1, 1, –7) is given by :
- 4x – 4y + z + 7 = 0
- 4x + 7y + 2z + 11 = 0
- 2x + y – z = 0
- 2x + y – 3z = 0
Q. The equation to the plane through the points (2, -1, 0), (3, -4, 5) parallel to a line with direction cosines proportional to (2, 3, 4 ) is 9x – 2y – 3z = k, where k is
- 10
- -10
- 20
- -20
Q. The equation of the plane passing through the points (2, –4, 0), (3, –4, 5) and parallel to the line 2x = 3y = 4z is :
- 29x – 27y – 22z = 85
- 73x + 61y – 22z = 85
- 125x – 90y – 79z = 340
- 32x – 21y – 36z = 85
Q.
The distance of the point (1, 3, -7) from the plane passing through the point (1, -1, -1) having normal perpendicular to both the lines
x−11=y+2−2=z−43 and x−22=y+1−1=z+7−1, is
20√74 units
10√83 units
5√83 units
10√74 units
Q.
Consider the lines
L1:x−12=y−1=z+31, L2:x−41=y+31=z+32, and the planes P1:7x+y+2z=3, P2:3x+5y−6z=4. Let ax+by+cz=d the equation of the plane passing through the point of intersection of lines L1 and L2 and perpendicular to planes P1 and P2.
Match List I with List II and select the correct answer using the code given below the lists.
ListIListIIPa=113Qb=2−3Rc=31Sd=4−2
P-3, Q-2, R-4, S-1
P-2, Q-4, R-1, S-3
P-1, Q-3, R-4, S-2
P-3, Q-2, R-1, S-4
Q. If the point (2, α, β) lies on the plane which passes through the point (3, 4, 2) and(7, 0, 6) and is perpendicular to the plane 2x−5y=15, then 2α−3β is equal to :
- 17
- 12
- 5
- 7