Key Concepts and Formulas

• Linear Combination of Coplanar Vectors: If vector $\overrightarrow u$ is coplanar with non-collinear vectors $\overrightarrow a$ and $\overrightarrow b$, then $\overrightarrow u$ can be expressed as $\overrightarrow u = x\overrightarrow a + y\overrightarrow b$ for some scalars $x$ and $y$.
• Dot Product Properties:
  • $\overrightarrow v \cdot \overrightarrow w = \overrightarrow w \cdot \overrightarrow v$ (Commutativity)
  • $\overrightarrow v \cdot \overrightarrow v = {\left| {\overrightarrow v } \right|^2}$
  • $\overrightarrow v \cdot (\overrightarrow w_1 + \overrightarrow w_2) = \overrightarrow v \cdot \overrightarrow w_1 + \overrightarrow v \cdot \overrightarrow w_2$ (Distributivity)
  • $(c\overrightarrow v) \cdot \overrightarrow w = c(\overrightarrow v \cdot \overrightarrow w)$ (Scalar multiplication)
• Perpendicular Vectors: Two vectors $\overrightarrow v$ and $\overrightarrow w$ are perpendicular if and only if $\overrightarrow v \cdot \overrightarrow w = 0$.

Step-by-Step Solution

Step 1: Express $\overrightarrow u$ as a linear combination of $\overrightarrow a$ and $\overrightarrow b$. Since $\overrightarrow u$ is coplanar with $\overrightarrow a = 2\widehat i + 3\widehat j - \widehat k$ and $\overrightarrow b = \widehat j + \widehat k$, we can write $\overrightarrow u$ as a linear combination of $\overrightarrow a$ and $\overrightarrow b$: $\overrightarrow u = x\overrightarrow a + y\overrightarrow b$ where $x$ and $y$ are scalar coefficients.

Step 2: Calculate necessary dot products and squared magnitudes. To simplify calculations later, we compute the dot products and squared magnitudes of the given vectors:

• ${\left| {\overrightarrow a } \right|^2} = \overrightarrow a \cdot \overrightarrow a = (2)^2 + (3)^2 + (-1)^2 = 4 + 9 + 1 = 14$
• ${\left| {\overrightarrow b } \right|^2} = \overrightarrow b \cdot \overrightarrow b = (0)^2 + (1)^2 + (1)^2 = 0 + 1 + 1 = 2$
• $\overrightarrow a \cdot \overrightarrow b = (2)(0) + (3)(1) + (-1)(1) = 0 + 3 - 1 = 2$

Step 3: Use the condition $\overrightarrow u \perp \overrightarrow a$ to form an equation. We are given that $\overrightarrow u$ is perpendicular to $\overrightarrow a$. This means their dot product is zero: $\overrightarrow u \cdot \overrightarrow a = 0$ Substituting $\overrightarrow u = x\overrightarrow a + y\overrightarrow b$: $(x\overrightarrow a + y\overrightarrow b) \cdot \overrightarrow a = 0$ Using the distributive property of the dot product: $x(\overrightarrow a \cdot \overrightarrow a) + y(\overrightarrow b \cdot \overrightarrow a) = 0$ Substituting the pre-calculated values: $x(14) + y(2) = 0$ Dividing by 2, we get our first equation: $7x + y = 0 \quad \ldots(1)$

Step 4: Use the condition $\overrightarrow u \cdot \overrightarrow b = 24$ to form another equation. We are also given that $\overrightarrow u \cdot \overrightarrow b = 24$. Substituting $\overrightarrow u = x\overrightarrow a + y\overrightarrow b$: $(x\overrightarrow a + y\overrightarrow b) \cdot \overrightarrow b = 24$ Using the distributive property of the dot product: $x(\overrightarrow a \cdot \overrightarrow b) + y(\overrightarrow b \cdot \overrightarrow b) = 24$ Substituting the pre-calculated values: $x(2) + y(2) = 24$ Dividing by 2, we get our second equation: $x + y = 12 \quad \ldots(2)$

Step 5: Solve the system of linear equations for $x$ and $y$. We have the following system of equations:

1. $7x + y = 0$
2. $x + y = 12$

From equation (1), we can express $y$ as $y = -7x$. Substitute this into equation (2): $x + (-7x) = 12$ $-6x = 12$ $x = -2$ Now substitute the value of $x$ back into the expression for $y$: $y = -7(-2) = 14$ So, we have $x = -2$ and $y = 14$. This means $\overrightarrow u = -2\overrightarrow a + 14\overrightarrow b$.

Step 6: Calculate ${\left| {\overrightarrow u } \right|^2}$. The magnitude squared of $\overrightarrow u$ is given by $\overrightarrow u \cdot \overrightarrow u$. ${\left| {\overrightarrow u } \right|^2} = \overrightarrow u \cdot \overrightarrow u$ Substitute $\overrightarrow u = x\overrightarrow a + y\overrightarrow b$: ${\left| {\overrightarrow u } \right|^2} = (x\overrightarrow a + y\overrightarrow b) \cdot \overrightarrow u$ Using the distributive property: ${\left| {\overrightarrow u } \right|^2} = x(\overrightarrow a \cdot \overrightarrow u) + y(\overrightarrow b \cdot \overrightarrow u)$ From Step 3, we know $\overrightarrow u \cdot \overrightarrow a = 0$, so $\overrightarrow a \cdot \overrightarrow u = 0$. From the problem statement (and Step 4), we know $\overrightarrow u \cdot \overrightarrow b = 24$, so $\overrightarrow b \cdot \overrightarrow u = 24$. Substituting these values and the determined values of $x$ and $y$: ${\left| {\overrightarrow u } \right|^2} = (-2)(0) + (14)(24)$ ${\left| {\overrightarrow u } \right|^2} = 0 + 336$ ${\left| {\overrightarrow u } \right|^2} = 336$

Common Mistakes & Tips

• Expanding $\overrightarrow u$ first: Avoid substituting $x$ and $y$ back into $\overrightarrow u = x\overrightarrow a + y\overrightarrow b$ to find the component form of $\overrightarrow u$ and then calculating its magnitude. This is often more calculation-intensive and prone to arithmetic errors. The method of using $\overrightarrow u \cdot \overrightarrow u = x(\overrightarrow a \cdot \overrightarrow u) + y(\overrightarrow b \cdot \overrightarrow u)$ is more efficient as it directly uses the given conditions.
• Sign errors: Be careful with signs when solving the system of linear equations or when calculating dot products.
• Vector Coplanarity: Remember that if $\overrightarrow u$ is coplanar with $\overrightarrow a$ and $\overrightarrow b$, and $\overrightarrow a$ and $\overrightarrow b$ are non-collinear, then $\overrightarrow u$ can be uniquely represented as $x\overrightarrow a + y\overrightarrow b$.

Summary

We began by expressing the vector $\overrightarrow u$, which is coplanar with $\overrightarrow a$ and $\overrightarrow b$, as a linear combination $\overrightarrow u = x\overrightarrow a + y\overrightarrow b$. We then used the given conditions $\overrightarrow u \cdot \overrightarrow a = 0$ and $\overrightarrow u \cdot \overrightarrow b = 24$ to form a system of two linear equations in $x$ and $y$. Solving this system yielded $x = -2$ and $y = 14$. Finally, we calculated ${\left| {\overrightarrow u } \right|^2}$ efficiently by using the identity ${\left| {\overrightarrow u } \right|^2} = \overrightarrow u \cdot \overrightarrow u = x(\overrightarrow a \cdot \overrightarrow u) + y(\overrightarrow b \cdot \overrightarrow u)$, substituting the known values of the dot products and the determined coefficients.

The final answer is $\boxed{336}$.