1. Key Concepts and Formulas

• Normal Vector of a Plane: For a plane given by the equation $Ax + By + Cz + D = 0$, its normal vector is $\vec{n} = A\widehat{i} + B\widehat{j} + C\widehat{k}$. This vector is perpendicular to the plane.
• Direction Vector of the Line of Intersection of Two Planes: If a line $L$ is the intersection of two planes with normal vectors $\vec{n_1}$ and $\vec{n_2}$, then the direction vector $\vec{b}$ of the line $L$ is perpendicular to both $\vec{n_1}$ and $\vec{n_2}$. Therefore, $\vec{b}$ is parallel to their cross product: $\vec{b} = \vec{n_1} \times \vec{n_2}$.
• Angle Between a Line and a Plane: The angle $\theta$ between a line with direction vector $\vec{b}$ and a plane with normal vector $\vec{n}$ is given by the formula: $\sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$ This formula is used because if $\alpha$ is the angle between the direction vector of the line and the normal vector of the plane, then $\theta = 90^\circ - \alpha$. Thus, $\sin \theta = \sin(90^\circ - \alpha) = \cos \alpha = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$.

2. Step-by-Step Solution

Step 1: Identify the Normal Vector of the Given Plane

• What we are doing: We need to find the normal vector of the plane $x + y + z = 5$. This vector will be used in the angle formula.
• Why: The normal vector is a fundamental property of a plane, indicating its orientation in space. It is essential for calculating the angle between the plane and the line.
• Calculation: The given plane is $P_1: x + y + z = 5$. Comparing this to the general form $Ax + By + Cz + D = 0$, we find $A=1$, $B=1$, and $C=1$.
• Result: The normal vector to the plane $P_1$ is $\vec{n_1} = \widehat{i} + \widehat{j} + \widehat{k}$.
• Magnitude: The magnitude of this normal vector is $|\vec{n_1}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$.

Step 2: Determine the Direction Vector of the Line of Intersection

• What we are doing: The line $L$ is the intersection of two planes, $P_2: 3x + 4y + z - 1 = 0$ and $P_3: 5x + 8y + 2z + 14 = 0$. We need to find the direction vector $\vec{b}$ of this line.
• Why: The direction vector of the line is crucial for applying the angle formula. Since the line lies in both planes, it must be perpendicular to the normal vectors of both planes. Therefore, its direction is given by the cross product of the normal vectors of these two planes.
• Calculation:
  • First, find the normal vectors of $P_2$ and $P_3$:
    • For $P_2: 3x + 4y + z - 1 = 0$, the normal vector is $\vec{n_2} = 3\widehat{i} + 4\widehat{j} + \widehat{k}$.
    • For $P_3: 5x + 8y + 2z + 14 = 0$, the normal vector is $\vec{n_3} = 5\widehat{i} + 8\widehat{j} + 2\widehat{k}$.
  • Next, calculate the cross product $\vec{n_2} \times \vec{n_3}$ to find the direction vector $\vec{b}$ of the line $L$: $\vec{b} = \vec{n_2} \times \vec{n_3} = \begin{vmatrix} \widehat{i} & \widehat{j} & \widehat{k} \\ 3 & 4 & 1 \\ 5 & 8 & 2 \end{vmatrix}$ $\vec{b} = \widehat{i}(4 \cdot 2 - 1 \cdot 8) - \widehat{j}(3 \cdot 2 - 1 \cdot 5) + \widehat{k}(3 \cdot 8 - 4 \cdot 5)$ $\vec{b} = \widehat{i}(8 - 8) - \widehat{j}(6 - 5) + \widehat{k}(24 - 20)$ $\vec{b} = 0\widehat{i} - 1\widehat{j} + 4\widehat{k} = -\widehat{j} + 4\widehat{k}$
• Result: The direction vector of the line of intersection is $\vec{b} = -\widehat{j} + 4\widehat{k}$.
• Magnitude: The magnitude of this direction vector is $|\vec{b}| = \sqrt{0^2 + (-1)^2 + 4^2} = \sqrt{1 + 16} = \sqrt{17}$.

Step 3: Calculate the Dot Product of the Line's Direction Vector and the Plane's Normal Vector

• What we are doing: We need to find the dot product $\vec{b} \cdot \vec{n_1}$. This is the numerator of our angle formula.
• Why: The dot product measures the projection of one vector onto another and is directly used in the formula for the angle between the line and the plane.
• Calculation:
  • $\vec{b} = -\widehat{j} + 4\widehat{k}$
  • $\vec{n_1} = \widehat{i} + \widehat{j} + \widehat{k}$
  • $\vec{b} \cdot \vec{n_1} = (0)(1) + (-1)(1) + (4)(1) = 0 - 1 + 4 = 3$
• Result: The dot product $\vec{b} \cdot \vec{n_1} = 3$.

Step 4: Apply the Formula for the Angle Between a Line and a Plane

• What we are doing: Substitute the calculated values into the formula $\sin \theta = \frac{|\vec{b} \cdot \vec{n_1}|}{|\vec{b}| |\vec{n_1}|}$ to find the angle $\theta$.
• Why: This is the final step to directly compute the required angle using the derived vectors.
• Calculation:
  • $|\vec{b} \cdot \vec{n_1}| = |3| = 3$
  • $|\vec{b}| = \sqrt{17}$
  • $|\vec{n_1}| = \sqrt{3}$
  • $\sin \theta = \frac{3}{\sqrt{17} \cdot \sqrt{3}} = \frac{3}{\sqrt{51}}$
  • To simplify, we can rationalize the numerator by multiplying by $\frac{\sqrt{3}}{\sqrt{3}}$: $\sin \theta = \frac{3}{\sqrt{51}} = \frac{3}{\sqrt{3} \cdot \sqrt{17}} = \frac{\sqrt{3} \cdot \sqrt{3}}{\sqrt{3} \cdot \sqrt{17}} = \frac{\sqrt{3}}{\sqrt{17}}$
  • This can also be written as: $\sin \theta = \sqrt{\frac{3}{17}}$
• Result: The angle $\theta$ is $\sin^{-1}\left(\sqrt{\frac{3}{17}}\right)$.

3. Common Mistakes & Tips

• Confusing Sine and Cosine: A common error is to use $\cos \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$ for the angle between a line and a plane. This formula gives the angle between the line and the normal to the plane, not the plane itself. Always remember to use $\sin \theta$ for the angle between a line and a plane.
• Cross Product Calculation Errors: Be meticulous when calculating the cross product of the normal vectors to find the direction vector of the line. A single sign error or incorrect determinant expansion will lead to a wrong direction vector.
• Absolute Value: Don't forget the absolute value in the numerator of the formula, $|\vec{b} \cdot \vec{n}|$, as the angle is conventionally taken as acute, hence its sine must be positive.

4. Summary

To find the angle between a plane and the line of intersection of two other planes, we first determined the normal vector of the given plane. Then, we found the direction vector of the line of intersection by taking the cross product of the normal vectors of the two planes defining the line. Finally, we used the formula $\sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$ to calculate the angle, where $\vec{b}$ is the direction vector of the line and $\vec{n}$ is the normal vector of the plane. The calculated angle was $\sin^{-1}\left(\sqrt{\frac{3}{17}}\right)$.

5. Final Answer

The angle between the plane $x + y + z = 5$ and the line of intersection of the planes $3x + 4y + z - 1 = 0$ and $5x + 8y + 2z + 14 = 0$ is $\sin^{-1}\left(\sqrt{\frac{3}{17}}\right)$. The final answer is \boxed{{\sin ^{ - 1}}\left( {\sqrt {{\raise0.5ex\hbox{$\scriptstyle 3$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle {17}$}}} } \right)}, which corresponds to option (A).