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

The Correct Option is A

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)