Key Concepts and Formulas

• 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 derived from the fact that the angle $\theta$ between the line and the plane is complementary to the angle $\phi$ between the line and the plane's normal vector (i.e., $\theta = 90^\circ - \phi$), and thus $\sin \theta = \cos \phi$.
• Direction Vector of a Line: For a line expressed in symmetric form $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$, its direction vector is $\vec{b} = a\hat{i} + b\hat{j} + c\hat{k}$.
• Normal Vector of a Plane: For a plane expressed in general form $Ax + By + Cz + D = 0$, its normal vector is $\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$.

Step-by-Step Solution

Step 1: Identify the Direction Vector of the Line

We begin by extracting the direction vector from the given equation of the line. The line is given by: $\frac{x + 1}{1} = \frac{y - 1}{2} = \frac{z - 2}{2}$ Comparing this with the standard symmetric form $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$, we can identify the direction ratios $(a, b, c)$ as $(1, 2, 2)$. Therefore, the direction vector of the line is: $\vec{b} = 1\hat{i} + 2\hat{j} + 2\hat{k}$

Step 2: Identify the Normal Vector of the Plane

Next, we extract the normal vector from the given equation of the plane. The plane is given by: $2x - y + \sqrt{\lambda}z + 4 = 0$ Comparing this with the standard general form $Ax + By + Cz + D = 0$, we identify the coefficients of $x, y, z$ as $A=2$, $B=-1$, and $C=\sqrt{\lambda}$. Therefore, the normal vector to the plane is: $\vec{n} = 2\hat{i} - 1\hat{j} + \sqrt{\lambda}\hat{k}$ Note: For $\sqrt{\lambda}$ to be a real number, it is implicitly assumed that $\lambda \ge 0$. This will be important for simplifying absolute values later.

Step 3: Calculate the Dot Product of the Vectors

Now, we compute the dot product of the direction vector of the line ($\vec{b}$) and the normal vector of the plane ($\vec{n}$). The dot product is the sum of the products of their corresponding components: $\vec{b} \cdot \vec{n} = (1)(2) + (2)(-1) + (2)(\sqrt{\lambda})$ $\vec{b} \cdot \vec{n} = 2 - 2 + 2\sqrt{\lambda}$ $\vec{b} \cdot \vec{n} = 2\sqrt{\lambda}$

Step 4: Calculate the Magnitudes of the Vectors

We need the magnitudes (lengths) of both vectors for the formula. The magnitude of a vector $p\hat{i} + q\hat{j} + r\hat{k}$ is $\sqrt{p^2 + q^2 + r^2}$.

• Magnitude of $\vec{b}$: $|\vec{b}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$

• Magnitude of $\vec{n}$: $|\vec{n}| = \sqrt{2^2 + (-1)^2 + (\sqrt{\lambda})^2} = \sqrt{4 + 1 + \lambda} = \sqrt{5 + \lambda}$

Step 5: Substitute into the Formula and Solve for $\lambda$

We are given that the angle $\theta$ between the line and the plane satisfies $\sin \theta = \frac{1}{3}$. We substitute the calculated values into the formula $\sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$: $\frac{1}{3} = \frac{|2\sqrt{\lambda}|}{3 \cdot \sqrt{5 + \lambda}}$ Since we established that $\lambda \ge 0$, $2\sqrt{\lambda}$ will always be non-negative, so $|2\sqrt{\lambda}| = 2\sqrt{\lambda}$. $\frac{1}{3} = \frac{2\sqrt{\lambda}}{3\sqrt{5 + \lambda}}$ We can cancel the '3' from both denominators: $1 = \frac{2\sqrt{\lambda}}{\sqrt{5 + \lambda}}$ Rearrange the equation to isolate the square roots: $\sqrt{5 + \lambda} = 2\sqrt{\lambda}$ To eliminate the square roots and solve for $\lambda$, we square both sides of the equation: $(\sqrt{5 + \lambda})^2 = (2\sqrt{\lambda})^2$ $5 + \lambda = 4\lambda$ Now, we solve for $\lambda$: $5 = 4\lambda - \lambda$ $5 = 3\lambda$ $\lambda = \frac{5}{3}$

Step 6: Verify the Solution

It is a good practice to verify solutions, especially after squaring both sides of an equation, to ensure no extraneous solutions have been introduced. Substitute $\lambda = \frac{5}{3}$ back into the equation before squaring, i.e., $\sqrt{5 + \lambda} = 2\sqrt{\lambda}$: $\sqrt{5 + \frac{5}{3}} = 2\sqrt{\frac{5}{3}}$ $\sqrt{\frac{15+5}{3}} = 2\sqrt{\frac{5}{3}}$ $\sqrt{\frac{20}{3}} = 2\sqrt{\frac{5}{3}}$ We know that $\sqrt{\frac{20}{3}} = \sqrt{4 \cdot \frac{5}{3}} = \sqrt{4} \cdot \sqrt{\frac{5}{3}} = 2\sqrt{\frac{5}{3}}$. $2\sqrt{\frac{5}{3}} = 2\sqrt{\frac{5}{3}}$ The solution is consistent. Also, $\lambda = 5/3$ satisfies the condition $\lambda \ge 0$.

Common Mistakes & Tips

• Angle Confusion: Be careful not to confuse the angle $\theta$ between the line and the plane with the angle $\phi$ between the line and the normal to the plane. The formula $\cos \phi = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$ is for $\phi$, whereas $\sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$ is for $\theta$. Remember that $\theta = 90^\circ - \phi$.
• Absolute Value: Always include the absolute value in the numerator, $|\vec{b} \cdot \vec{n}|$, as the angle $\theta$ between a line and a plane is conventionally taken as an acute angle ($0 \le \theta \le 90^\circ$), for which $\sin \theta \ge 0$.
• Verification after Squaring: When you square both sides of an equation to remove square roots, always check your final answer in the original equation (or the equation just before squaring) to ensure it is a valid solution and not an extraneous one.

Summary

This problem involved finding an unknown parameter $\lambda$ by utilizing the formula for the angle between a line and a plane. We systematically extracted the direction vector of the line and the normal vector of the plane from their given equations. We then computed their dot product and magnitudes, substituting these values into the $\sin \theta$ formula along with the given value of $\sin \theta$. Solving the resulting algebraic equation, which involved squaring both sides, yielded the value of $\lambda$. Finally, the solution was verified to ensure its validity.

The final answer is $\boxed{{5 \over 3}}$, which corresponds to option (A).