# Three Dimensional Geometry - Online Test

Q1. Find the equation of the line in cartesian form that passes through the point (– 2, 4, – 5) and parallel to the line given by
Explaination / Solution:

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 .
Therefore ,
.

Q2. Find the cartesian equations of the lines that passes through the origin and (5, – 2, 3).
Explaination / Solution:

The coordinates of the origin are ( 0 , 0 ,0 ) , therefore , , and l = 5 , m = - 2 and n = 3 ,
therefore the equation in Cartesian form is given by:

i.e. .

Q3. Find the vector equations of the line that passes through the points (3, – 2, – 5), (3, – 2, 6).
Explaination / Solution:

Here ,  and
Therefore , the vector equation is :

i.e..

Q4.
Find the values of p so that the linesare at right angles.

Explaination / Solution:

Q5. Find the shortest distance between the lines
Explaination / Solution:

Q6. Find the shortest distance between the lines
Explaination / Solution:

Q7. Find the shortest distance between the lines :
Explaination / Solution:

On comparing the given equations with:
, we get:

Q8. Find the shortest distance between the lines
Explaination / Solution:

The given equations can be reduced as:

Q9. In the vector form, equation of a plane which is at a distance d from the origin, and  is the unit vector normal to the plane through the origin is
Explaination / Solution:

In the vector form, equation of a plane which is at a distance d from the origin, and  is the unit vector normal to the plane through the origin is given by :

Q10. 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.
Explaination / Solution:

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.