Question

Equation of the plane passing through the origin and perpendicular to the planes \( x + 2y - z = 1 \) and \( 3x - 4y + z = 5 \) is

• A. \( x + 2y - 5z = 0 \)
• B. \( x - 2y + 5z = 0 \)
• C. \( x + 2y + 5z = 0 \)
• D. \( 3x + y - 5z = 0 \)

Hint:

A plane perpendicular to two given planes has a normal vector that is the cross product of the normals of the two planes.

Answer 1

The Correct Option is C

Step 1: Let normal vectors of the given planes be: For \( x + 2y - z = 1 \), normal vector is \( \vec{n}_1 = \langle 1, 2, -1 \rangle \) For \( 3x - 4y + z = 5 \), normal vector is \( \vec{n}_2 = \langle 3, -4, 1 \rangle \) Step 2: The required plane is perpendicular to both given planes. So, its normal vector is the cross product: \[ \vec{n} = \vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -1 \\ 3 & -4 & 1 \end{vmatrix} = \hat{i}(2 . 1 - (-1) . (-4)) - \hat{j}(1 . 1 - (-1) . 3) + \hat{k}(1 . -4 - 2 . 3) \] \[ = \hat{i}(2 - 4) - \hat{j}(1 - (-3)) + \hat{k}(-4 - 6) = \langle -2, -4, -10 \rangle \] Step 3: The plane passes through origin and has this normal vector, so equation is: \[ -2x - 4y - 10z = 0 \Rightarrow x + 2y + 5z = 0 \]