Answer:
Listed below:
Step-by-step explanation:
P: 192
A:2079
B. Area
C. Perimeter
Hope this helps :)
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The required plane Π contains the line
L: (-1,1,2)+t(7,6,2)
means that Π is perpendicular to the direction vector of the line L, namely
vl=<7,6,2>
It is also required that Π be perpendicular to the plane
Π 1 : 5y-7z+8=0
means that Π is also perpendicular to the normal vector of the given plane, vp=<0,5,-7>.
Thus the normal vector of the required plane, Π can be obtained by the cross product of vl and vp, or vl x vp:
i j k
7 6 2
0 5 -7
=<-42-10, 0+49, 35-0>
=<-52, 49, 35>
which is the normal vector of Π
Since Π has to contain the line, it must pass through the point (-1,1,2), so the equation of the plane is
Π : -52(x-(-1))+49(y-1)+35(z-2)=0
=>
Π : -52x+49y+35z = 171
Check that normal vector of plane is orthogonal to line direction vector
<-52,49,35>.<7,6,2>
=-364+294+70
=0 ok
