Question

Find an equation for the plane that contains the line v = (−1, 1, 2) + t(7, 6, 2) and is perpendicular to the plane 8x + 5y − 7z + 8 = 0.

Answer 1

The equation for the plane that contains the given line and is perpendicular to the provided plane is 8x + 5y - 7z + 17 = 0.

Step-by-step explanation:

To find an equation for the plane that contains the line given by v = (−1, 1, 2) + t(7, 6, 2) and is perpendicular to the plane described by 8x + 5y − 7z + 8 = 0, we first need to understand that the normal vector of this plane will be parallel to the normal vector (8, 5, -7) of the given plane since both planes are perpendicular to each other.

Knowing that the line lies on the required plane, we can use the point (−1, 1, 2) that the line passes through, along with the parallel normal vector, to write the equation of our plane. The general form of the equation of a plane is Ax + By + Cz + D = 0, where (A, B, C) is the normal vector to the plane. Substituting our point into the equation, we calculate the constant D.

Thus, the equation of our plane will be 8x + 5y − 7z + D = 0. Inserting the point (−1, 1, 2) into this equation gives us 8(-1) + 5(1) - 7(2) + D = 0, which simplifies to -8 + 5 - 14 + D = 0. Solving for D yields D = 17. Hence, the final equation of the plane is 8x + 5y − 7z + 17 = 0.

Answer 2

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 &Pi; be perpendicular to the plane
&Pi; 1 : 5y-7z+8=0
means that &Pi; is also perpendicular to the normal vector of the given plane, vp=<0,5,-7>.

Thus the normal vector of the required plane, &Pi; 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 &Pi;

Since &Pi; has to contain the line, it must pass through the point (-1,1,2), so the equation of the plane is
&Pi; : -52(x-(-1))+49(y-1)+35(z-2)=0
=>
&Pi; : -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