Find the equation of the plane that contains the line x = 1 + t , y = 2 − t , z = 4 − 3 t and is parallel to the plane 5 x + 2 y + z = 1So far I figured the direction vector of the line is &lt; 1 , − 1 , − 3 &gt; and that the symmetric equation of the line is x − 1 = 2 − y = z − 4 − 3 . But I am lost on how to go from here.

Question

Find the equation of the plane that contains the line   x  =  1  +  t  ,  y  =  2  −  t  ,  z  =  4  −  3  t and is parallel to the plane   5  x  +  2  y  +  z  =  1So far I figured the direction vector of the line is   &amp;lt;  1  ,  −  1  ,  −  3  &amp;gt; and that the symmetric equation of the line is   x  −  1  =  2  −  y  =            z      −      4              −      3      . But I am lost on how to go from here.

Nicholas Mathis · Accepted Answer

The plane P you&#039;re looking for is parallel to the one given by   5  x  +  2  y  +  z  =  1. Thus P has an equation of the form                    (1)                    5        x        +        2        y        +        z        =        α            where   α is a constant that we need to determine. That&#039;s because both planes are orthogonal to vector (5,2,1).To do that, remember P to contain that one line. So any point on that line should be on the plane. Let&#039;s pick the point   C  =  (  1  ,  2  ,  4  ), which is definitely on that line (I found it by taking t=0).Let&#039;s plug the coordinates of C inside (1):  5  ×  1  +  2  ×  2  +  1  ×  4  =  α  =  13Thus P has equation  5  x  +  2  y  +  z  =  13

Find the equation of the plane that contains the line x=1+t,y=2−t,z=4−3t

Answered question

Answer & Explanation

New Questions in Precalculus