Test: Three Dimensional Geometry- 2 - JEE MCQ

By definition : The Skew lines are lines in different planes which are neither parallel nor intersecting .

If a line has the direction ratios – 18, 12, – 4, then its direction cosines are given by:

In Cartesian co – ordinate system Equation of a plane which is at a distance d from the origin and the direction cosines of the normal to the plane are l, m, n is given by : lx + my + nz = d.

Angle between skew lines is the angle between two intersecting lines drawn from any point 

The equation of the line which passes through the point (1, 2, 3) and is parallel to the vector  Let vector   and vector 

In vector form The equation of a plane through a point whose position vector is  perpendicular to the vector  Is given by : 

We have, 4x + 8y + z – 8 = 0 and y + z – 4 = 0. Let be the angle between the planes, then 

If l₁, m₁, n₁ and l₂, m₂, n₂ are the direction cosines of two lines; and θ is the acute angle between the two lines; then the cosine of the angle between these two lines is given by : 

, then , its Cartesian equation is given by :

In cartesian co – ordinate system : Equation of a plane passing through three non collinear points (x₁, y₁, z₁),(x₂, y₂, z₂) and (x₃, y₃, z₃) is given  

As we know that the length of the perpendicular from point 
P(x₁,y₁,z₁) from the plane a₁x+b₁y+c₁z+d₁ = 0 is given by: 

If a₁, b₁, c₁ and a₂, b₂, c₂ are the direction ratios of two lines and θθ is the acute angle between the two lines; then , the cosine of the angle between these two lines is given by :

Find the equation of the line in cartesian form that passes through the point (– 2, 4, – 5) and parallel to the line given by

is given by:
 And l = 3 , m = 5 and n = 6 .

Vector equation of a plane that contains three non collinear points having position vectors

Let 
be the position vector of the point  here,
. Therefore, the required vector equation of the plane is: 

As we know that the length of the perpendicular from point P(x₁,y₁,z₁) from the plane

Here, P(3, - 2,1) is the point and equation of Plane is 2x - y + 2z+3 = 0

Therefore, the perpendicular distance is :

 

Vector equation of a line that passes through the given point whose position vector is and parallel to a given vector is given by : 

Give lines are :
 and 
The D.R's of the lines are -3, 2p/7, 2 and -3p/7, 1, -5

In vector form:
Vector equation of a plane that passes through the intersection of planes
 expressed in terms of a non – zero constant λ is given by :

In cartesian co-ordinate system :
Equation of a plane passing through three non collinear

Points (x₁, y₁, z₁) , (x₂, y₂, z₂) and (x₃, y₃, z₃) is given by :



Therefore, the equations of the planes that passes through three points (1,1,0), (1,2,1),  (-2,2,-1) is given by :



⇒ (x-1)(-2) - (y-1) (3) + 3z = 0
⇒ 2x+3y - 3z = 5

As we know that the length of the perpendicular from point P(x₁,y₁,z₁) from the plane


Here, P(2,3,-5) is the point and equation of plane is x+2y - 2z = 9

Therefore, the perpendicular distance is :