Key Concepts and Formulas

1. Normal Vector to a Plane: A plane passing through three non-collinear points $A, B, C$ has a normal vector $\vec{n}$ that is perpendicular to any two non-parallel vectors lying in the plane. A common way to find $\vec{n}$ is by taking the cross product of two such vectors, for instance, $\vec{n} = \overrightarrow{AB} \times \overrightarrow{AC}$.
2. Magnitude of a Vector: For a vector $\vec{v} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$, its magnitude is $|\vec{v}| = \sqrt{x^2 + y^2 + z^2}$.
3. Dot Product of Vectors: For two vectors $\vec{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}$ and $\vec{b} = b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k}$, their dot product is $\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3$.
4. Projection of a Vector onto a Plane: The length of the projection of a vector $\vec{v}$ onto a plane with normal vector $\vec{n}$ can be found using the Pythagorean theorem. If $L_{\text{plane}}$ is the length of the projection onto the plane and $L_n$ is the length of the scalar projection of $\vec{v}$ onto the normal vector $\vec{n}$, then:
   • $L_n = \frac{|\vec{v} \cdot \vec{n}|}{|\vec{n}|}$ (This is the component of $\vec{v}$ perpendicular to the plane).
   • $L_{\text{plane}} = \sqrt{|\vec{v}|^2 - L_n^2}$ (This is derived from a right triangle where $|\vec{v}|$ is the hypotenuse, $L_n$ is one leg, and $L_{\text{plane}}$ is the other leg).

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Step-by-Step Solution

Step 1: Determine the Normal Vector to the Plane

To find the normal vector $\vec{n}$ of the plane, we first need two vectors lying within the plane. We can form these vectors using the given points A(1, 2, 3), B(2, 3, 1), and C(2, 4, 2).

• Calculate $\overrightarrow{AB}$: This vector goes from point A to point B. $\overrightarrow{AB} = B - A = (2-1)\mathbf{i} + (3-2)\mathbf{j} + (1-3)\mathbf{k} = \mathbf{i} + \mathbf{j} - 2\mathbf{k}$
• Calculate $\overrightarrow{AC}$: This vector goes from point A to point C. $\overrightarrow{AC} = C - A = (2-1)\mathbf{i} + (4-2)\mathbf{j} + (2-3)\mathbf{k} = \mathbf{i} + 2\mathbf{j} - \mathbf{k}$
• Calculate the normal vector $\vec{n}$: The normal vector is perpendicular to both $\overrightarrow{AB}$ and $\overrightarrow{AC}$, so we find it using their cross product. $\vec{n} = \overrightarrow{AB} \times \overrightarrow{AC} = \left| {\begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & -2 \\ 1 & 2 & -1 \end{matrix}} \right|$ Expanding the determinant: $\vec{n} = \mathbf{i}((1)(-1) - (-2)(2)) - \mathbf{j}((1)(-1) - (-2)(1)) + \mathbf{k}((1)(2) - (1)(1))$ $\vec{n} = \mathbf{i}(-1 + 4) - \mathbf{j}(-1 + 2) + \mathbf{k}(2 - 1)$ $\vec{n} = 3\mathbf{i} - \mathbf{j} + \mathbf{k}$
• Calculate the magnitude of the normal vector $|\vec{n}|$: This will be needed for the projection formula. $|\vec{n}| = \sqrt{3^2 + (-1)^2 + 1^2} = \sqrt{9 + 1 + 1} = \sqrt{11}$

Step 2: Identify the Vector to be Projected

The question asks for the projection of $\overrightarrow{OP}$ on the plane. Given O is the origin (0, 0, 0) and P is (2, -1, 1).

• Calculate $\overrightarrow{OP}$: $\overrightarrow{OP} = P - O = (2-0)\mathbf{i} + (-1-0)\mathbf{j} + (1-0)\mathbf{k} = 2\mathbf{i} - \mathbf{j} + \mathbf{k}$
• Calculate the magnitude of $\overrightarrow{OP}$: This is the length of the vector itself, which is the hypotenuse in our projection triangle. $|\overrightarrow{OP}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$

Step 3: Calculate the Length of the Projection onto the Plane

We will use the formula $L_{\text{plane}} = \sqrt{|\overrightarrow{OP}|^2 - L_n^2}$, where $L_n = \frac{|\overrightarrow{OP} \cdot \vec{n}|}{|\vec{n}|}$.

• Calculate the dot product $\overrightarrow{OP} \cdot \vec{n}$: This product helps determine the component of $\overrightarrow{OP}$ along the normal vector. $\overrightarrow{OP} \cdot \vec{n} = (2\mathbf{i} - \mathbf{j} + \mathbf{k}) \cdot (3\mathbf{i} - \mathbf{j} + \mathbf{k})$ $\overrightarrow{OP} \cdot \vec{n} = (2)(3) + (-1)(-1) + (1)(1) = 6 + 1 + 1 = 8$
• Calculate the scalar projection $L_n$ of $\overrightarrow{OP}$ onto $\vec{n}$: This is the length of the component of $\overrightarrow{OP}$ that is perpendicular to the plane. $L_n = \frac{|\overrightarrow{OP} \cdot \vec{n}|}{|\vec{n}|} = \frac{|8|}{\sqrt{11}} = \frac{8}{\sqrt{11}}$
• Calculate the length of the projection onto the plane $L_{\text{plane}}$: Now we apply the Pythagorean theorem. $L_{\text{plane}} = \sqrt{|\overrightarrow{OP}|^2 - L_n^2}$ Substitute the calculated values: $L_{\text{plane}} = \sqrt{(\sqrt{6})^2 - \left(\frac{8}{\sqrt{11}}\right)^2}$ $L_{\text{plane}} = \sqrt{6 - \frac{64}{11}}$ To simplify, find a common denominator: $L_{\text{plane}} = \sqrt{\frac{6 \times 11}{11} - \frac{64}{11}} = \sqrt{\frac{66 - 64}{11}}$ $L_{\text{plane}} = \sqrt{\frac{2}{11}}$

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Common Mistakes & Tips

• Cross Product Accuracy: Be extremely careful with signs and calculations when computing the cross product. A single error here will lead to an incorrect normal vector and subsequently, an incorrect final answer.
• Geometric Understanding: Always visualize the problem. The projection of a vector onto a plane forms a right-angled triangle with the original vector as the hypotenuse and the scalar projection onto the normal as one leg. This intuition helps in correctly applying the Pythagorean theorem.
• Magnitude vs. Scalar Projection: Distinguish between the magnitude of the vector $|\vec{v}|$ and its scalar projection onto the normal vector $L_n$. These are distinct quantities crucial for the formula.

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Summary

To find the length of the projection of $\overrightarrow{OP}$ onto the plane, we first determined the normal vector to the plane by taking the cross product of two vectors lying within it ($\overrightarrow{AB}$ and $\overrightarrow{AC}$). Then, we calculated the magnitude of the vector $\overrightarrow{OP}$ and the magnitude of the normal vector. Using these values, along with the dot product $\overrightarrow{OP} \cdot \vec{n}$, we found the scalar projection of $\overrightarrow{OP}$ onto the normal. Finally, applying the Pythagorean theorem, which relates the original vector, its component perpendicular to the plane, and its component parallel to the plane, we calculated the required projection length.

The final answer is $\boxed{\sqrt{\frac{2}{11}}}$, which corresponds to option (A).