Key Concepts and Formulas

• Magnitude of a vector: For a vector $\vec{v} = x\hat{i} + y\hat{j} + z\hat{k}$, its magnitude is $|\vec{v}| = \sqrt{x^2 + y^2 + z^2}$.
• Magnitude of the cross product: $|\vec{b} \times \vec{c}| = |\vec{b}||\vec{c}|\sin\theta$, where $\theta$ is the angle between $\vec{b}$ and $\vec{c}$.
• Magnitude of the sum of two vectors squared: $|\vec{b} + \vec{c}|^2 = |\vec{b}|^2 + |\vec{c}|^2 + 2\vec{b} \cdot \vec{c}$.
• Dot product: $\vec{b} \cdot \vec{c} = |\vec{b}||\vec{c}|\cos\theta$.
• Trigonometric identity: $\sin^2\theta + \cos^2\theta = 1$.

Step-by-Step Solution

Step 1: Calculate the magnitudes of the given vectors $\vec{a}$ and $\vec{c}$, and use the given $|\vec{b}|^2$. We are given $\vec{a}=3 \hat{i}+\hat{j}-\hat{k}$ and $\vec{c}=2 \hat{i}-3 \hat{j}+3 \hat{k}$. We are also given $|\vec{b}|^{2}=50$.

The magnitude of $\vec{a}$ is: $|\vec{a}| = \sqrt{(3)^2 + (1)^2 + (-1)^2} = \sqrt{9 + 1 + 1} = \sqrt{11}$ The magnitude of $\vec{c}$ is: $|\vec{c}| = \sqrt{(2)^2 + (-3)^2 + (3)^2} = \sqrt{4 + 9 + 9} = \sqrt{22}$ We are given $|\vec{b}|^2 = 50$.

Step 2: Use the cross product relationship to find $\sin\theta$. We are given $\vec{a} = \vec{b} \times \vec{c}$. Using the magnitude of the cross product formula, $|\vec{a}| = |\vec{b}||\vec{c}|\sin\theta$, where $\theta$ is the angle between $\vec{b}$ and $\vec{c}$.

Substituting the calculated magnitudes: $\sqrt{11} = \sqrt{50} \cdot \sqrt{22} \cdot \sin\theta$ $\sqrt{11} = \sqrt{50 \cdot 22} \cdot \sin\theta$ $\sqrt{11} = \sqrt{1100} \cdot \sin\theta$ $\sqrt{11} = \sqrt{100 \cdot 11} \cdot \sin\theta$ $\sqrt{11} = 10\sqrt{11} \cdot \sin\theta$ Solving for $\sin\theta$: $\sin\theta = \frac{\sqrt{11}}{10\sqrt{11}} = \frac{1}{10}$

Step 3: Find $\cos\theta$ using the trigonometric identity $\sin^2\theta + \cos^2\theta = 1$. From Step 2, we have $\sin\theta = \frac{1}{10}$. $\cos^2\theta = 1 - \sin^2\theta = 1 - \left(\frac{1}{10}\right)^2 = 1 - \frac{1}{100} = \frac{99}{100}$ Therefore, $\cos\theta = \pm\sqrt{\frac{99}{100}} = \pm\frac{\sqrt{99}}{10} = \pm\frac{3\sqrt{11}}{10}$. We have two possible values for $\cos\theta$.

Step 4: Calculate $|\vec{b} + \vec{c}|^2$ using the formula $|\vec{b} + \vec{c}|^2 = |\vec{b}|^2 + |\vec{c}|^2 + 2\vec{b} \cdot \vec{c}$. We know $|\vec{b}|^2 = 50$ and $|\vec{c}|^2 = 22$. The dot product $\vec{b} \cdot \vec{c} = |\vec{b}||\vec{c}|\cos\theta$. First, calculate $|\vec{b}||\vec{c}|$: $|\vec{b}||\vec{c}| = \sqrt{50} \cdot \sqrt{22} = \sqrt{1100} = 10\sqrt{11}$ Now, we consider the two cases for $\cos\theta$:

Case 1: $\cos\theta = \frac{3\sqrt{11}}{10}$ $|\vec{b} + \vec{c}|^2 = 50 + 22 + 2 \left(10\sqrt{11}\right) \left(\frac{3\sqrt{11}}{10}\right)$ $|\vec{b} + \vec{c}|^2 = 72 + 2 \cdot (3\sqrt{11} \cdot \sqrt{11})$ $|\vec{b} + \vec{c}|^2 = 72 + 2 \cdot (3 \cdot 11)$ $|\vec{b} + \vec{c}|^2 = 72 + 66 = 138$

Case 2: $\cos\theta = -\frac{3\sqrt{11}}{10}$ $|\vec{b} + \vec{c}|^2 = 50 + 22 + 2 \left(10\sqrt{11}\right) \left(-\frac{3\sqrt{11}}{10}\right)$ $|\vec{b} + \vec{c}|^2 = 72 - 2 \cdot (3\sqrt{11} \cdot \sqrt{11})$ $|\vec{b} + \vec{c}|^2 = 72 - 2 \cdot (3 \cdot 11)$ $|\vec{b} + \vec{c}|^2 = 72 - 66 = 6$

Step 5: Calculate the final expression $|72 - |\vec{b}+\vec{c}|^{2}|$. We need to evaluate $|72 - |\vec{b}+\vec{c}|^{2}|$.

Using the result from Case 1: $|72 - 138| = |-66| = 66$

Using the result from Case 2: $|72 - 6| = |66| = 66$ In both cases, the result is 66.

Common Mistakes & Tips

• Sign of $\cos\theta$: Remember that $\sin\theta$ being positive allows for both an acute ($\cos\theta > 0$) and an obtuse ($\cos\theta < 0$) angle. Both possibilities must be considered, but the final absolute value often resolves the ambiguity.
• Algebraic Simplification: Be careful with simplifying square roots and products involving them, especially when calculating the dot product term.
• Absolute Value at the End: Ensure the absolute value is applied to the entire expression $72 - |\vec{b}+\vec{c}|^{2}$, not just to the $|\vec{b}+\vec{c}|^{2}$ term.

Summary The problem requires a systematic application of vector properties. We first calculated the magnitudes of the given vectors. Then, using the cross product relationship, we determined the value of $\sin\theta$. The identity $\sin^2\theta + \cos^2\theta = 1$ was used to find the possible values of $\cos\theta$. Finally, we computed $|\vec{b}+\vec{c}|^2$ for both possible values of $\cos\theta$ and substituted these into the expression $|72 - |\vec{b}+\vec{c}|^{2}|$. The absolute value ensured a unique answer.

The final answer is $\boxed{66}$.