Question
Find the vector equation in scalar product form of the plane that contains the lines.
and 
Solution
The equation of lines are
...(1)
and
...(2)
Both the lines (1) and (2) pass through the point (1, 1, 0) whose position vector is
.
Hence both the lines are intersecting and are coplanar.
The equation of any plane containing the line (1) is
A (x - 1) + B (y - 1) + C (z - 0) = 0 ...(3)
where A+ 2 B - C = 0 ...(4)
Now plane is also normal to the vector
as plane is perp. to line (2).
∴ – A + B – 2 C = 0 ... (5)
Solving (4) and (5), we get,
or
or
Putting these values of A, B, C in (3), we get, (x - 1) - (y – 1) – (z – 0) = 0 or x – y – z = 0
This can be written as
or
...(1)and
...(2)Both the lines (1) and (2) pass through the point (1, 1, 0) whose position vector is
.Hence both the lines are intersecting and are coplanar.
The equation of any plane containing the line (1) is
A (x - 1) + B (y - 1) + C (z - 0) = 0 ...(3)
where A+ 2 B - C = 0 ...(4)
Now plane is also normal to the vector
as plane is perp. to line (2).∴ – A + B – 2 C = 0 ... (5)
Solving (4) and (5), we get,
or
or
Putting these values of A, B, C in (3), we get, (x - 1) - (y – 1) – (z – 0) = 0 or x – y – z = 0
This can be written as
or
