The plane $2x - y + 3z + 5 = 0$ is rotated through $90^\circ$ about its line of intersection with the plane $x + y + z = 1$. The equation of the plane in the new position is:

Question

Let $P$ be a point in the plane of the vectors \[ \vec{AB}=3\hat{i}+\hat{j}-\hat{k} \quad\text{and}\quad \vec{AC}=\hat{i}-\hat{j}+3\hat{k} \] such that $P$ is equidistant from the lines $AB$ and $AC$. If $|\vec{AP}|=\frac{\sqrt5}{2}$, then the area of triangle $ABP$ is:

Answer 1

Step 1: Determine the new plane equation after a 90° rotation about the line of intersection; begin by finding the line's equation.

Step 2: The given planes are defined by $2x - y + 3z + 5 = 0$ and $x + y + z = 1$. The line of intersection is found by solving these equations simultaneously.

Step 3: Solve the system of equations, expressing two variables in terms of the third. This yields the parametric equations for the line of intersection.

Step 4: Apply the 90° rotation using a 3D rotation matrix. This matrix is derived from the line's axis and the rotation angle.

Step 5: Apply the rotation to the plane equation and simplify to obtain the new equation $3x + 9y + z = 17$.

The plane $2x - y + 3z + 5 = 0$ is rotated through $90^\circ$ about its line of intersection with the plane $x + y + z = 1$. The equation of the plane in the new position is:

Show Hint

The Correct Option is B

Solution and Explanation

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