Plane is an important topic in the Mathematics section in BITSAT exam. Practising this topic will increase your score overall and make your conceptual grip on BITSAT exam stronger.
This article gives you a full set of BITSAT PYQs for Plane with explanations for effective preparation. Practice of BITSAT Mathematics PYQs including Plane questions regularly will improve accuracy, speed, and confidence in the BITSAT 2026 exam.
Also Read
BITSAT PYQs for Plane with Solutions
1.
Let R be the relation "is congruent to" on the set of all triangles in a plane. Is R:- Reflexive only
- Symmetric only
- Symmetric and reflexive only
- Equivalence relation
2.
The point of intersection of the line $\frac{x-1}{3}=\frac{y+2}{4}=\frac{z-3}{-2}$ and plane $2 x-y+3 z-1= 0$ is.- (10, -10, 3)
- (10, 10, -3)
- (-10, 10, 3)
- none of these
3.
Let the foot of perpendicular from a point \( P(1,2,-1) \) to the straight line \( L : \frac{x}{1} = \frac{y}{0} = \frac{z}{-1} \) be \( N \). Let a line be drawn from \( P \) parallel to the plane \( x + y + 2z = 0 \) which meets \( L \) at point \( Q \). If \( \alpha \) is the acute angle between the lines \( PN \) and \( PQ \), then \( \cos \alpha \) is equal to:- \( \frac{1}{\sqrt{5}} \)
- \( \frac{\sqrt{3}}{2} \)
- \( \frac{1}{\sqrt{3}} \)
- \( \frac{1}{2\sqrt{3}} \)
4.
Let the acute angle bisector of the two planes \( x - 2y - 2z + 1 = 0 \) and \( 2x - 3y - 6z + 1 = 0 \) be the plane \( P \). Then which of the following points lies on \( P \)?- \( (3, 1, -\frac{1}{2}) \)
- \( (-2, 0, -\frac{1}{2}) \)
- \( (0, 2, -4) \)
- \( (4, 0, -2) \)
5.
The equation of plane passing through a point $A (2,-1,3)$ and parallel to the vectors $a =(3,0,-1)$ and $b =(-3,2,2)$ is:- 2x - 3y + 6z - 25 = 0
- 2x - 3y + 6z + 25 = 0
- 3x - 2y + 6z - 25 = 0
- 3x - 2y + 6z + 25 = 0
Comments