Question

The plane that is perpendicular to the planes $x-y+2z-4=0$ and $2x-2y+z=0$ and passes through $(1,-2,1)$ is

• A. $x+y+1=0$
• B. $2x+y+z-1=0$
• C. $x+y+z=0$
• D. $2x+y-z+1=0$
• E. $x+z-2=0$

Answer 1

Answer: (A)

$x+y+1=0$

Answer 2

Given that:
The desired equation of the plane that is perpendicular to the given planes and passes through the point $(1, -2, 1)$

Then we can find,

The normal vector of a plane with equation $Ax + By + Cz + D = 0$ is $(A, B, C)$.

For the plane$x - y + 2z - 4 = 0$, the normal vector is $(1, -1, 2).$

For the plane $2x - 2y + z = 0$, the normal vector is $(2, -2, 1).$

Now, since our desired plane is perpendicular to both of these planes, its normal vector must be perpendicular to both $(1, -1, 2)$ and $(2, -2, 1)$. We can find the normal vector of the desired plane by taking the cross product of the normal vectors of the given planes:

Normal vector of the desired plane = $(1, -1, 2) × (2, -2, 1)$

To find the cross product, we can use the determinant method:

$i((-1×1)-(2×-2))-j((1×1)-(2×2))+k((1×-2)-(-1×2))$

$=3i+3j+0k$

So, the normal vector of the desired plane is (3, 3, 0).

Now, we have the normal vector of the desired plane (3, 3, 0) and a point it passes through (1, -2, 1). We can use this information to find the equation of the plane using the point-normal form of the plane equation:

$(x - x_0)a + (y - y_0)b + (z - z_0)c = 0$

where $(x_0, y_0, z_0)$ is the point the plane passes through, and $(a, b, c)$ is the normal vector.

Substitute the values:

$(x - 1)*3 + (y - (-2))*3 + (z - 1)*0 = 0$

Simplify:

$3x-3+3y+6=0$

$x+y+1=0$ , is the desired equation. (_Ans)