Three-dimensional Geometry Test - 11

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  a₁ x+b₁ y+c₁ z+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 :