Question

hi... maths doubt... answer only if you know

Answer 1

This is the question from 3 dimensional geometry,
So, equation of line is cartesian form is as follows,
$\frac{x-x1}{a} = \frac{y-y1}{b} = \frac{z-z1}{c} = λ$
where, x1, y1 and z1 are the points and a,b and c are the direction ratio of the vector parallel to this line.
The general point lying on this line is given by,
[tex]x = aλ + x1, \\ y = bλ + y1, \\ z = cλ + z1.[/tex]

Now come to the question,
The general point on line L1,
$x = 4λ1 + 0, \\ y = 6λ1 - 1, \\ z = 2λ1 - 1$             [As a,b and c can be                                                                                             calculated by                                                                                                                    vector                                                                                                                    B - A ]

Similarly, for line L2,
$x = -7λ2 + 3, \\ y = -5λ2 + 9, \\ z = 0λ2 + 4$

For intersection of line they must be meet at one point (x,y,z).
General point of line L1 = General point of Line 2         [Compare x, y and z                                                                                        coordinate differently.] 
$4λ1 + 0 = -7λ2 + 3 ....(1), \\ 6λ1 - 1 = -5λ2 + 9 ....(2), \\ 2λ1 - 1 = 0λ2 + 4 ....(3)$

So, we have three equations,
Determine the value of λ1 and λ2 from taking any two equations and check the intersection by putting value of λ1 and λ2 in third equations.
On solving for λ1 and λ2, we get
λ1 = 5/2, λ2= -1 
Now, putting the value of λ1 and λ2 in third equation.
2λ1 - 1 = 4.

On putting,
4 = 4       (L.H.S. = R.H.S.)
Therefore, these two lines intersects.

Intersecting point = (4λ1,6λ1 - 1, 2λ1 - 1)
Put the value of λ1,
Intersecting point = ( 10,14,4).