SOLUTION: Find the equation of the line joining (1,3,-1) to (2,4,7). Show that it is parallel to the plane 2x+6y-z=5

Algebra ->  Linear-equations -> SOLUTION: Find the equation of the line joining (1,3,-1) to (2,4,7). Show that it is parallel to the plane 2x+6y-z=5      Log On


   



Question 984077: Find the equation of the line joining (1,3,-1) to (2,4,7). Show that it is parallel to the plane 2x+6y-z=5
Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
Instead of doing your problem for you, I will instead do one exactly like
yours step by step so you can use it as a model to do yours by:

Find the equation of the line joining (2,4,-3) to (3,9,6). Show that it is
parallel to the plane x-2y+z=7

A parametric equation for a line through the point



parallel to the vector

matrix%281%2C7%2C%22%3C%22%2Ca%2C%22%2C%22%2Cb%2C%22%2C%22%2Cc%2C%22%3E%22%29

is:



A symmetric equation for the line is

%28x-x%5B1%5D%29%2Fa%22%22=%22%22%28y-y%5B1%5D%29%2Fb%22%22=%22%22%28z-z%5B1%5D%29%2Fc

-----------------------------

To find a vector parallel to the line, 

matrix%281%2C7%2C%22%3C%22%2Ca%2C%22%2C%22%2Cb%2C%22%2C%22%2Cc%2C%22%3E%22%29

we subtract coordinates:

%22%22=%22%22matrix%281%2C7%2C%22%3C%22%2C1%2C%22%2C%22%2C5%2C%22%2C%22%2C9%2C%22%3E%22%29

A parametric equation for the line is then:



matrix%281%2C5%2Cx=2%2B1t%2C+%22%2C%22%2C++y=4%2B5t%2C+%22%2C%22%2C+z=-3%2B9t%29

A symmetric equation for the line is

%28x-x%5B1%5D%29%2Fa%22%22=%22%22%28y-y%5B1%5D%29%2Fb%22%22=%22%22%28z-z%5B1%5D%29%2Fc

%28x-2%29%2F1%22%22=%22%22%28y-4%29%2F5%22%22=%22%22%28z-%28-3%29%29%2F9

%28x-2%29%2F1%22%22=%22%22%28y-4%29%2F5%22%22=%22%22%28z%2B3%29%2F9

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Given the equation of the plane  x-2y+z=7, a normal to the plane is
a vector whose components are the coefficients of x,y and z in the
equation of the plane.  So

matrix%281%2C7%2C%22%3C%22%2C1%2C%22%2C%22%2C-2%2C%22%2C%22%2C1%2C%22%3E%22%29  is a vector normal to the plane 
x+2y-z=7.

So we merely have to show that matrix%281%2C7%2C%22%3C%22%2C1%2C%22%2C%22%2C-2%2C%22%2C%22%2C1%2C%22%3E%22%29 is
perpendicular to the vector parallel to the line, which is matrix%281%2C7%2C%22%3C%22%2C1%2C%22%2C%22%2C5%2C%22%2C%22%2C9%2C%22%3E%22%29   

To show two vectors are perpendicular we show that their dot product is 0.

matrix%281%2C7%2C%22%3C%22%2C1%2C%22%2C%22%2C-2%2C%22%2C%22%2C1%2C%22%3E%22%29matrix%281%2C7%2C%22%3C%22%2C1%2C%22%2C%22%2C5%2C%22%2C%22%2C9%2C%22%3E%22%29%22%22=%22%22%281%29%281%29%2B%28-2%29%285%29%2B%281%29%289%29%22%22=%22%221-10%2B9%22%22=%22%220.

Now do yours exactly the same way.

Edwin