Find symmetric equations for the line of intersection of the planes.
$ z = 2x - y - 5 $ , $ z = 4x + 3y -5 $
the question they asked me to find the symmetry equations for the line of intersections of these planes, the number one is equal to two, X minus by minus five. And play number two is equal to four. Express three by minus five. Yeah, so the equations of the planes can be re adding some regular formats. Two X minus one, man is equal to five. And second plane is four X plus three by men. Is that equal to five? So the normal vectors of each of these planes and one vector is equal to 2 -1 -1 and and two vector is equal to 4, 3 and -1. And so we have to find the perpendicular of the line of intersection of these planes. So and vector is equal to anyone victor across into victor. And according to the cross production we find the value of n vector is equal to four, icap minus two, Jacob plus 10-K cup. Therefore Normal vector to the line of intersection of this plane is equal to 4 -2 and 10. Yeah. Okay. Yeah. So after finding the normal victor to the point of intersection on this plane, we have to find out point of intersection of these planes coordinates. Mhm. So in order to find the coordinates we have to put an arbitrary value To any variable, let zero equal to zero. Therefore the first equation becomes two weeks minus y equal to five. And the second equation becomes four X last three way Is equal to five. So if we calculate the values of X and y. From this equations we multiply the first equation by three we get uh the coefficients for x will be six X -3, Y is equal to 15. Therefore, if we add district questions, why 10 will get cancelled and the question will become 10, X is equal to 20. Therefore from here X is equal to two. So from the above equation, if we put the value of X, we can get the value of Why? So four into 2 plus three, Y equal to five. From the second equation, we can find the value of Y. So three Y is equal to five minus eight. Therefore Why is equal to -1. We have found out coordinates of the point of intersection of these two planes. So the coordinates are mhm Okay, Mhm. 2 -1 and zero. Now we have to find this symmetric equation at the line of intersection of these planes that can be evaluated as follows. So, mm hmm. So for finding this imagery equation of the line of intersection of these two planes, we have found out the normal vector, the point of intersection of these two planes that is equal to 4 -2 in 10. And the point of coordinate at the point of intersection of these two planes is equal to 2 -1 and zero. And therefore the symmetric equation can originals x minus the point according it Or x coordinate -2 divided by the normal vector for this coordinate, that is four is equal to why minus the point of coordinate for the y axis, that is minus one divided by The normal vector corresponding to this coordinate. That is -2 is equal to zero access minus the point of coordinates, or this access divided by the corresponding normal victor Value, that is 10. So after simplifying the situation it is equal to x minus two by four is equal to Y plus one by minus two is equal to zed by 10. And hence this is the symmetric equation at the line of intersection of the given planes.