Key Concepts and Formulas

• Vector Representation in a Plane: A vector $\overrightarrow x$ lying in the plane spanned by two non-collinear vectors $\overrightarrow a$ and $\overrightarrow b$ can be uniquely expressed as a linear combination: $\overrightarrow x = c_1 \overrightarrow a + c_2 \overrightarrow b$, where $c_1$ and $c_2$ are scalars.
• Perpendicular Vectors: Two vectors are perpendicular if and only if their dot product is zero: $\overrightarrow u \cdot \overrightarrow v = 0$.
• Scalar Projection: The scalar projection of vector $\overrightarrow P$ onto vector $\overrightarrow Q$ is given by $\text{proj}_{\overrightarrow Q} \overrightarrow P = \frac{\overrightarrow P \cdot \overrightarrow Q}{|\overrightarrow Q|}$.
• Dot Product Properties: $\overrightarrow u \cdot \overrightarrow v = \overrightarrow v \cdot \overrightarrow u$, and $\overrightarrow u \cdot \overrightarrow u = |\overrightarrow u|^2$.
• Magnitude of a Vector: For $\overrightarrow v = v_x \widehat i + v_y \widehat j + v_z \widehat k$, $|\overrightarrow v|^2 = v_x^2 + v_y^2 + v_z^2$.

Step-by-Step Solution

Step 1: Represent $\overrightarrow x$ as a linear combination of $\overrightarrow a$ and $\overrightarrow b$. Since $\overrightarrow x$ lies in the plane containing $\overrightarrow a$ and $\overrightarrow b$, we can write: $\overrightarrow x = c_1 \overrightarrow a + c_2 \overrightarrow b$ Given: $\overrightarrow a = 2\widehat i - \widehat j + \widehat k$ $\overrightarrow b = \widehat i + 2\widehat j - \widehat k$ Substituting these into the expression for $\overrightarrow x$: $\overrightarrow x = c_1 (2\widehat i - \widehat j + \widehat k) + c_2 (\widehat i + 2\widehat j - \widehat k)$ $\overrightarrow x = (2c_1 + c_2)\widehat i + (-c_1 + 2c_2)\widehat j + (c_1 - c_2)\widehat k$

Step 2: Use the perpendicularity condition to form an equation. We are given that $\overrightarrow x$ is perpendicular to $\overrightarrow c = 3\widehat i + 2\widehat j - \widehat k$. This means their dot product is zero: $\overrightarrow x \cdot \overrightarrow c = 0$. We can express this in terms of $c_1$ and $c_2$: $(c_1 \overrightarrow a + c_2 \overrightarrow b) \cdot \overrightarrow c = 0$ $c_1 (\overrightarrow a \cdot \overrightarrow c) + c_2 (\overrightarrow b \cdot \overrightarrow c) = 0$ First, calculate the dot products $\overrightarrow a \cdot \overrightarrow c$ and $\overrightarrow b \cdot \overrightarrow c$: $\overrightarrow a \cdot \overrightarrow c = (2)(3) + (-1)(2) + (1)(-1) = 6 - 2 - 1 = 3$ $\overrightarrow b \cdot \overrightarrow c = (1)(3) + (2)(2) + (-1)(-1) = 3 + 4 + 1 = 8$ Substituting these values into the equation: $3c_1 + 8c_2 = 0 \quad \text{(Equation 1)}$

Step 3: Use the projection condition to form another equation. The projection of $\overrightarrow x$ on $\overrightarrow a$ is given as $\frac{17\sqrt 6}{2}$. The formula for scalar projection is $\text{proj}_{\overrightarrow a} \overrightarrow x = \frac{\overrightarrow x \cdot \overrightarrow a}{|\overrightarrow a|}$. First, find the magnitude of $\overrightarrow a$: $|\overrightarrow a| = \sqrt{(2)^2 + (-1)^2 + (1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$ Now, set up the projection equation: $\frac{\overrightarrow x \cdot \overrightarrow a}{|\overrightarrow a|} = \frac{17\sqrt 6}{2}$ $\frac{\overrightarrow x \cdot \overrightarrow a}{\sqrt{6}} = \frac{17\sqrt 6}{2}$ Solving for $\overrightarrow x \cdot \overrightarrow a$: $\overrightarrow x \cdot \overrightarrow a = \frac{17\sqrt 6}{2} \cdot \sqrt{6} = \frac{17 \cdot 6}{2} = 17 \cdot 3 = 51$ Now, express $\overrightarrow x \cdot \overrightarrow a$ in terms of $c_1$ and $c_2$: $\overrightarrow x \cdot \overrightarrow a = (c_1 \overrightarrow a + c_2 \overrightarrow b) \cdot \overrightarrow a = c_1 (\overrightarrow a \cdot \overrightarrow a) + c_2 (\overrightarrow b \cdot \overrightarrow a)$ We know $\overrightarrow a \cdot \overrightarrow a = |\overrightarrow a|^2 = (\sqrt{6})^2 = 6$. And $\overrightarrow b \cdot \overrightarrow a = (1)(2) + (2)(-1) + (-1)(1) = 2 - 2 - 1 = -1$. Substituting these values: $\overrightarrow x \cdot \overrightarrow a = c_1 (6) + c_2 (-1) = 6c_1 - c_2$ Equating this to the value of $\overrightarrow x \cdot \overrightarrow a$: $6c_1 - c_2 = 51 \quad \text{(Equation 2)}$

Step 4: Solve the system of linear equations for $c_1$ and $c_2$. We have the system:

1. $3c_1 + 8c_2 = 0$
2. $6c_1 - c_2 = 51$ From Equation 1, $3c_1 = -8c_2$, so $c_1 = -\frac{8}{3}c_2$. Substitute this into Equation 2: $6\left(-\frac{8}{3}c_2\right) - c_2 = 51$ $-16c_2 - c_2 = 51$ $-17c_2 = 51$ $c_2 = \frac{51}{-17} = -3$ Now, substitute $c_2 = -3$ back into the expression for $c_1$: $c_1 = -\frac{8}{3}(-3) = 8$ Thus, $c_1 = 8$ and $c_2 = -3$.

Step 5: Calculate $|\overrightarrow x|^2$. We have $\overrightarrow x = c_1 \overrightarrow a + c_2 \overrightarrow b = 8\overrightarrow a - 3\overrightarrow b$. $\overrightarrow x = 8(2\widehat i - \widehat j + \widehat k) - 3(\widehat i + 2\widehat j - \widehat k)$ $\overrightarrow x = (16\widehat i - 8\widehat j + 8\widehat k) - (3\widehat i + 6\widehat j - 3\widehat k)$ $\overrightarrow x = (16 - 3)\widehat i + (-8 - 6)\widehat j + (8 - (-3))\widehat k$ $\overrightarrow x = 13\widehat i - 14\widehat j + 11\widehat k$ Now, calculate the squared magnitude of $\overrightarrow x$: $|\overrightarrow x|^2 = (13)^2 + (-14)^2 + (11)^2$ $|\overrightarrow x|^2 = 169 + 196 + 121$ $|\overrightarrow x|^2 = 486$

Common Mistakes & Tips

• Ensure that $\overrightarrow a$ and $\overrightarrow b$ are not collinear before assuming they span a plane. In this case, they are clearly not scalar multiples.
• Carefully perform dot product calculations, paying close attention to signs, as errors here propagate.
• Distinguish between scalar projection (a scalar value) and vector projection (a vector). The problem provides a scalar value.

Summary The problem required finding the squared magnitude of a vector $\overrightarrow x$ constrained to a plane and satisfying perpendicularity and projection conditions. We represented $\overrightarrow x$ as a linear combination of the basis vectors of the plane, $\overrightarrow x = c_1 \overrightarrow a + c_2 \overrightarrow b$. The given conditions were translated into two linear equations involving $c_1$ and $c_2$. Solving this system yielded the values of $c_1$ and $c_2$. Finally, we constructed the vector $\overrightarrow x$ and computed its squared magnitude.

The final answer is $\boxed{486}$.