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Problem 40 Hard Difficulty

Find an equation of the plane.

The plane that passes through the line of intersection of the planes $ x - z = 1 $ and $ y + 2z = 3 $ and is perpendicular to the plane $ x + y - 2z = 1 $

Answer

$x+y+z-4=0$

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Video Transcript

in the question they're asking to find the equation of the plane that passes to the line of intersection of the plains given by the equations X minus that equal to one and wipe list to that equal to three. And the plane is also for particular to another plane which has equation expressed by minus two that equal to one. So in order to find the equation of the plane asking the question first we have to find out the normal vector to the plane. So the two equations of the planes that are intersecting is ex miners that equal to one and why plus two? Z equal to three. So from the first equation we can rearrange in terms of Z that is equal to x minus one. And therefore putting this value of zero. In the second equation we get y plus two into x minus one is equal to three. Therefore y plus two, X minus two is equal to three. And from here we find the equation as why plus two X is equal to five. And therefore from this equation, okay we assume that if by equal to one then two X equal to four, therefore X is equal to two. And from above equation we know that Z equal to x minus one. That is here that equal to one. Therefore the coordinates after putting by equal to one is point A which lies in the point of intersection of the two planes is having the coordinates two 11. And in order to find the normal victor of the plane, we have to find another point. So assuming why? Equal to, can assume any. Mhm. Mhm. Any value in order to find another point. So as you mean that X equal to zero. We get by equal to five and they're equal to minus one. There's another point B is 05 minus one. So this is a point A. And this is another point B. From these two points we can find the direction vector of this line. That is a. B vector is equal to zero minus two. I kept plus five minus one, Jacob plus minus one minus one kicker. And this is equal to minus two, four minus two. So this is a normal victor of the plane. And let it be equal to be one victor. Another normal vector to the plane can be found out from the perpendicular to the plane. Who's the equation is given us? So equation of the perpendicular plane to the given plane is explains why minus two is equal to one. From here we can find another normal vector to the plane. That is video cap is equal to one comma, one comma minus two. Therefore, in order to find a normal victor to the required plane, we have to conclude the cross product. So end cap is equal to be one cap. Cross, we took up this is equal to we had read the table for the cross product. Okay and put the values of the one victory in which a vector that is minus 24 minus two and 11 minus two. After calculating the value of the cross product, we get the value of the normal victor to the plane as minus one minus one and minus one. Therefore aggression of the plane from normal victor to the plane and the point E. Mhm. That is 211. We get the equation of the plane as minus one x minus two minus one, y minus one minus one, zed minus one equal to zero. Therefore, after solving this equation, the equation finally becomes X plus Y plus said minus for equal to zero. And this is the required equation of the plane asking the question.