Key Concepts and Formulas

• Dot Product: For vectors $\vec{u} = u_1\widehat{i} + u_2\widehat{j} + u_3\widehat{k}$ and $\vec{v} = v_1\widehat{i} + v_2\widehat{j} + v_3\widehat{k}$, their dot product is $\vec{u} \cdot \vec{v} = u_1v_1 + u_2v_2 + u_3v_3$.
• Perpendicular Vectors: Two non-zero vectors are perpendicular if and only if their dot product is zero ($\vec{u} \cdot \vec{v} = 0$).
• Magnitude of a Vector: For a vector $\vec{v} = v_1\widehat{i} + v_2\widehat{j} + v_3\widehat{k}$, its squared magnitude is $|\vec{v}|^2 = v_1^2 + v_2^2 + v_3^2$.
• Lagrange's Identity: For any two vectors $\vec{u}$ and $\vec{v}$, $|\vec{u} \times \vec{v}|^2 = |\vec{u}|^2 |\vec{v}|^2 - (\vec{u} \cdot \vec{v})^2$.

Step-by-Step Solution

Step 1: Use the perpendicularity condition between $\vec{a}$ and $\vec{b}$

We are given that $\vec{a}$ is perpendicular to $\vec{b}$. This means their dot product is zero. $\vec{a} = \widehat{i} + 5\widehat{j} + \alpha \widehat{k}$ $\vec{b} = \widehat{i} + 3\widehat{j} + \beta \widehat{k}$ $\vec{a} \cdot \vec{b} = (1)(1) + (5)(3) + (\alpha)(\beta) = 0$ $1 + 15 + \alpha\beta = 0$ $16 + \alpha\beta = 0$ $\alpha\beta = -16 \quad (*)$ This equation establishes a relationship between $\alpha$ and $\beta$.

Step 2: Use the magnitude of the cross product condition for $\vec{b}$ and $\vec{c}$

We are given $|\vec{b} \times \vec{c}| = 5\sqrt{3}$. Squaring both sides gives $|\vec{b} \times \vec{c}|^2 = (5\sqrt{3})^2 = 75$. We will use Lagrange's Identity: $|\vec{b} \times \vec{c}|^2 = |\vec{b}|^2 |\vec{c}|^2 - (\vec{b} \cdot \vec{c})^2$. First, let's compute the necessary magnitudes and dot product: $\vec{b} = \widehat{i} + 3\widehat{j} + \beta \widehat{k} \implies |\vec{b}|^2 = 1^2 + 3^2 + \beta^2 = 1 + 9 + \beta^2 = 10 + \beta^2$ $\vec{c} = - \widehat{i} + 2\widehat{j} - 3\widehat{k} \implies |\vec{c}|^2 = (-1)^2 + 2^2 + (-3)^2 = 1 + 4 + 9 = 14$ $\vec{b} \cdot \vec{c} = (1)(-1) + (3)(2) + (\beta)(-3) = -1 + 6 - 3\beta = 5 - 3\beta$ Now substitute these into Lagrange's Identity: $75 = (10 + \beta^2)(14) - (5 - 3\beta)^2$ $75 = 140 + 14\beta^2 - (25 - 30\beta + 9\beta^2)$ $75 = 140 + 14\beta^2 - 25 + 30\beta - 9\beta^2$ $75 = 5\beta^2 + 30\beta + 115$ Rearrange into a quadratic equation: $5\beta^2 + 30\beta + 115 - 75 = 0$ $5\beta^2 + 30\beta + 40 = 0$ Divide by 5: $\beta^2 + 6\beta + 8 = 0$ Factor the quadratic equation: $(\beta + 2)(\beta + 4) = 0$ This yields two possible values for $\beta$: $\beta = -2 \quad \text{or} \quad \beta = -4$

Step 3: Calculate $|\vec{a}|^2$ for each possible value of $\beta$

We need to find the greatest value of $|\vec{a}|^2$. The squared magnitude of $\vec{a}$ is: $|\vec{a}|^2 = 1^2 + 5^2 + \alpha^2 = 1 + 25 + \alpha^2 = 26 + \alpha^2$ We use the relationship $\alpha\beta = -16$ from Step 1 to find $\alpha$ for each value of $\beta$.

Case 1: $\beta = -2$ From $\alpha\beta = -16$, we have $\alpha(-2) = -16$, which gives $\alpha = 8$. Then, $|\vec{a}|^2 = 26 + \alpha^2 = 26 + 8^2 = 26 + 64 = 90$.

Case 2: $\beta = -4$ From $\alpha\beta = -16$, we have $\alpha(-4) = -16$, which gives $\alpha = 4$. Then, $|\vec{a}|^2 = 26 + \alpha^2 = 26 + 4^2 = 26 + 16 = 42$.

Comparing the two values, 90 and 42, the greatest value of $|\vec{a}|^2$ is 90.

Common Mistakes & Tips

• Be careful with algebraic manipulations, especially when squaring binomials involving variables (e.g., $(5 - 3\beta)^2$).
• Ensure that you consider all possible values for $\alpha$ and $\beta$ derived from the given conditions before determining the maximum value.
• Lagrange's Identity is often more efficient than directly computing the cross product and then its magnitude, especially when dealing with unknown components.

Summary

We began by using the condition that $\vec{a}$ is perpendicular to $\vec{b}$ to establish a relationship between $\alpha$ and $\beta$ ($\alpha\beta = -16$). Next, we applied Lagrange's Identity to the given magnitude of the cross product $|\vec{b} \times \vec{c}|$ to obtain a quadratic equation in $\beta$. Solving this quadratic equation yielded two possible values for $\beta$ ($\beta = -2$ and $\beta = -4$). For each of these values, we found the corresponding value of $\alpha$ and then calculated $|\vec{a}|^2 = 26 + \alpha^2$. By comparing the two resulting values for $|\vec{a}|^2$, we identified the greatest value.

The final answer is $\boxed{90}$.