How this equation represent the equation of line?

[Hyperspace2]

Hello people. In my textbooks the question is asked as following
1. Find the S.D(short distance)between the lines
a) x-2/3=y-6/9=z-9/5
b) 5x + 2y-6z-5 = 0 = 4x+6y+7z-8
My problem is not about of solution.

Solution is preety well given in book. But
My main question is
how equation b represent the equation of line


Further,it states

5x + 2y-6z-5 + K(4x+6y+7z-8) = 0 represents the equation of plane which consist line.
How could this be?

[tiny-tim]

Hello Hyperspace2! :smile:
b) 5x + 2y-6z-5 = 0 = 4x+6y+7z-8

My main question is
how equation b represent the equation of line

For example, x = y = 0 is obviously a line (the z-axis).

Similarly, any two independent linear equations make a line.

This is because any one linear equation makes a plane, so two together makes all the points on both planes, ie the intersection, which is a line! :wink:
5x + 2y-6z-5 + K(4x+6y+7z-8) = 0 represents the equation of plane which consist line.
How could this be?

It means that, for any fixed value of K, that is the equation of a plane which includes line b.

It's true because if you put any point on line b into that equation, it becomes 0 + K0 = 0 … which is obviously true! :smile:

[Hyperspace2]

For example, x = y = 0 is obviously a line (the z-axis).




.

This is because any one linear equation makes a plane, so two together makes all the points on both planes, ie the intersection, which is a line! :wink:


It means that, for any fixed value of K, that is the equation of a plane which includes line b. :
.
It's true because if you put any point on line b into that equation, it becomes 0 + K0 = 0 … which is obviously true! :smile:

Thanks I never thought of that . I am crystal clear now.

[HallsofIvy]

Another way of looking at it is this:
The "compound" equation (x-2)/3= (y-6)/9= (z-9)/5 (I presume you intended those parentheses) can be broken into two equations, (x- 2)/3= (y- 6)/9 and (y- 6)/9= (z- 9)/5 which are the equations of two planes. The two equations, together, are satisfied by points that lie on the intersection of the two planes, a line.

Finally, Since those three fractions are equal to one another, set them all equal to "t":
(x- 2)/3= t so x- 2= 3t and x= 3t+ 2,
(y- 6)/9= t so y- 6= 9t and y= 9t+ 6
(z- 9)/6= t so z- 9= 6t and z= 6t+ 9

Those are three linear parametric equations and so are parametric equations of a line.

[Hyperspace2]

Another way of looking at it is this:
The "compound" equation (x-2)/3= (y-6)/9= (z-9)/5 (I presume you intended those parentheses) can be broken into two equations, (x- 2)/3= (y- 6)/9 and (y- 6)/9= (z- 9)/5 which are the equations of two planes. The two equations, together, are satisfied by points that lie on the intersection of the two planes, a line.

Finally, Since those three fractions are equal to one another, set them all equal to "t":
(x- 2)/3= t so x- 2= 3t and x= 3t+ 2,
(y- 6)/9= t so y- 6= 9t and y= 9t+ 6
(z- 9)/6= t so z- 9= 6t and z= 6t+ 9

Those are three linear parametric equations and so are parametric equations of a line.

Thanks sir for making a more clear explanation.