Key Concepts and Formulas

1. Normal Vector of a Plane: For a plane given by the vector equation $\overrightarrow r \cdot \vec n = d$, the vector $\vec n$ is the normal vector to the plane. In Cartesian form, for $Ax + By + Cz = D$, the normal vector is $\vec n = A\widehat i + B\widehat j + C\widehat k$. The components of $\vec n$ are the direction ratios of the normal.
2. Normal Vector of a Plane through Three Points: If a plane passes through three non-collinear points $A, B, C$, then two vectors lying in the plane can be formed (e.g., $\overrightarrow{AB}$ and $\overrightarrow{AC}$). The cross product of these two vectors, $\overrightarrow{AB} \times \overrightarrow{AC}$, yields a vector perpendicular to the plane, which is its normal vector.
3. Direction Vector of the Line of Intersection of Two Planes: The line of intersection of two planes $P_1$ and $P_2$ is simultaneously perpendicular to the normal vector of $P_1$ ($\vec{n_1}$) and the normal vector of $P_2$ ($\vec{n_2}$). Therefore, the direction vector of this line of intersection is given by the cross product of their normal vectors: $\vec{d} = \vec{n_1} \times \vec{n_2}$. The components of this resultant vector are the direction ratios of the line.

Step-by-Step Solution

Step 1: Determine the Normal Vector of Plane $P_1$

The equation of plane $P_1$ is given as $\overrightarrow r \,.\,\left( {2\widehat i + \widehat j - 3\widehat k} \right) = 4$. Reasoning: This equation is in the standard vector form $\overrightarrow r \cdot \vec n = d$, where $\vec n$ is the normal vector to the plane. Comparing the given equation with the standard form, we can directly identify the normal vector for plane $P_1$. Let $\vec{n_1}$ be the normal vector to $P_1$. $\vec{n_1} = 2\widehat i + \widehat j - 3\widehat k$ The direction ratios of the normal to $P_1$ are $\langle 2, 1, -3 \rangle$.

Step 2: Determine the Normal Vector of Plane $P_2$

Plane $P_2$ passes through three points: $A(2, -3, 2)$, $B(2, -2, -3)$, and $C(1, -4, 2)$. Reasoning: To find the normal vector of a plane defined by three points, we form two vectors lying within the plane using these points. Their cross product will be perpendicular to both vectors, thus providing the normal vector to the plane.

Let's form two vectors within the plane using points A, B, and C: Vector $\overrightarrow{AB}$: $\overrightarrow{AB} = B - A = (2-2)\widehat i + (-2 - (-3))\widehat j + (-3-2)\widehat k$ $\overrightarrow{AB} = 0\widehat i + 1\widehat j - 5\widehat k$

Vector $\overrightarrow{BC}$: $\overrightarrow{BC} = C - B = (1-2)\widehat i + (-4 - (-2))\widehat j + (2 - (-3))\widehat k$ $\overrightarrow{BC} = -1\widehat i - 2\widehat j + 5\widehat k$

Now, let $\vec{n_2}$ be the normal vector to $P_2$. We find $\vec{n_2}$ by taking the cross product of $\overrightarrow{AB}$ and $\overrightarrow{BC}$: $\vec{n_2} = \overrightarrow{AB} \times \overrightarrow{BC} = \left| {\begin{matrix} {\widehat i} & {\widehat j} & {\widehat k} \\ 0 & 1 & { - 5} \\ { - 1} & { - 2} & 5 \end{matrix}} \right|$ Expanding the determinant: $\vec{n_2} = \widehat i ((1)(5) - (-5)(-2)) - \widehat j ((0)(5) - (-5)(-1)) + \widehat k ((0)(-2) - (1)(-1))$ $\vec{n_2} = \widehat i (5 - 10) - \widehat j (0 - 5) + \widehat k (0 + 1)$ $\vec{n_2} = -5\widehat i + 5\widehat j + 1\widehat k$ The direction ratios of the normal to $P_2$ are $\langle -5, 5, 1 \rangle$.

Step 3: Find the Direction Ratios of the Line of Intersection of $P_1$ and $P_2$

Reasoning: The line of intersection of planes $P_1$ and $P_2$ is perpendicular to both their normal vectors, $\vec{n_1}$ and $\vec{n_2}$. Therefore, its direction vector can be found by taking the cross product of $\vec{n_1}$ and $\vec{n_2}$.

We have $\vec{n_1} = 2\widehat i + 1\widehat j - 3\widehat k$ and $\vec{n_2} = -5\widehat i + 5\widehat j + 1\widehat k$. Let $\vec{d}$ be the direction vector of the line of intersection. $\vec{d} = \vec{n_1} \times \vec{n_2} = \left| {\begin{matrix} {\widehat i} & {\widehat j} & {\widehat k} \\ 2 & 1 & { - 3} \\ { - 5} & 5 & 1 \end{matrix}} \right|$ Expanding the determinant: $\vec{d} = \widehat i ((1)(1) - (-3)(5)) - \widehat j ((2)(1) - (-3)(-5)) + \widehat k ((2)(5) - (1)(-5))$ $\vec{d} = \widehat i (1 - (-15)) - \widehat j (2 - 15) + \widehat k (10 - (-5))$ $\vec{d} = \widehat i (1 + 15) - \widehat j (2 - 15) + \widehat k (10 + 5)$ $\vec{d} = 16\widehat i - (-13)\widehat j + 15\widehat k$ $\vec{d} = 16\widehat i + 13\widehat j + 15\widehat k$ The direction ratios of the line of intersection are $\langle 16, 13, 15 \rangle$.

Step 4: Identify $\alpha$ and $\beta$ and Calculate $\alpha + \beta$

We are given that the direction ratios of the line of intersection are $16, \alpha, \beta$. From our calculation, the direction ratios are $\langle 16, 13, 15 \rangle$. Comparing these components: $\alpha = 13$ $\beta = 15$ Now, we need to find the value of $\alpha + \beta$: $\alpha + \beta = 13 + 15$ $\alpha + \beta = 28$

Common Mistakes & Tips

• Cross Product Calculation: Be very careful with the signs and order of terms when calculating cross products. A common error is forgetting the negative sign for the $\widehat j$ component.
• Vector Formation: Ensure correct subtraction of coordinates when forming vectors from points (e.g., $\overrightarrow{AB} = B - A$).
• Geometric Interpretation: Always keep in mind the geometric meaning of normal vectors (perpendicular to the plane) and the direction vector of the line of intersection (perpendicular to both plane normals). This helps in choosing the correct vector operations.

Summary

To find the direction ratios of the line of intersection of two planes, we first determine the normal vector for each plane. For the first plane, the normal vector is directly given by its vector equation. For the second plane, the normal vector is found by taking the cross product of two vectors formed by the three given points on the plane. Finally, the direction vector of the line of intersection is obtained by taking the cross product of the two normal vectors. By comparing the components of this direction vector with the given form $(16, \alpha, \beta)$, we found $\alpha=13$ and $\beta=15$. The sum $\alpha+\beta$ is then $13+15=28$.

The final answer is $\boxed{1}$