Question

Let $\vec{a} = \hat{i} + 5\hat{j} + \alpha\hat{k}$, $\vec{b} = \hat{i} + 3\hat{j} + \beta\hat{k}$ and $\vec{c} = -\hat{i} + 2\hat{j} - 3\hat{k}$ be three vectors such that, $|\vec{b} \times \vec{c}| = 5\sqrt{3}$ and $\vec{a}$ is perpendicular to $\vec{b}$. Then the greatest amongst the values of $|\vec{a}|^2$ is _________.

Hint:

Magnitude squared calculation is straightforward: $|\vec{V}|^2 = V_x^2 + V_y^2 + V_z^2$. Perpendicularity always implies dot product equals zero.

Answer 1

Correct Answer: 90

Step 1: Understanding the Concept:
We use the vector cross product to find a condition for \( \beta \), and the dot product property for perpendicular vectors (\( \vec{a} \cdot \vec{b} = 0 \)) to find \( \alpha \). Finally, we calculate the magnitude of vector \( \vec{a} \).
Step 2: Detailed Explanation:
1. Compute \( \vec{b} \times \vec{c} \):

\[ \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 3 & \beta \\ -1 & 2 & -3 \end{vmatrix} = \hat{i}(-9 - 2\beta) - \hat{j}(-3 + \beta) + \hat{k}(2 + 3) = (-9 - 2\beta)\hat{i} + (3 - \beta)\hat{j} + 5\hat{k} \]

2. Given \( |\vec{b} \times \vec{c}| = 5\sqrt{3} \implies |\vec{b} \times \vec{c}|^2 = 75 \).

\[ (9 + 2\beta)^2 + (3 - \beta)^2 + 25 = 75 \]

\[ 81 + 36\beta + 4\beta^2 + 9 - 6\beta + \beta^2 + 25 = 75 \]

\[ 5\beta^2 + 30\beta + 115 = 75 \implies 5\beta^2 + 30\beta + 40 = 0 \implies \beta^2 + 6\beta + 8 = 0 \]

Roots are \( \beta = -2 \) and \( \beta = -4 \).
3. Given \( \vec{a} \perp \vec{b} \implies \vec{a} \cdot \vec{b} = 0 \).

\[ (\hat{i} + 5\hat{j} + \alpha\hat{k}) \cdot (\hat{i} + 3\hat{j} + \beta\hat{k}) = 1 + 15 + \alpha\beta = 16 + \alpha\beta = 0 \implies \alpha\beta = -16 \]

If \( \beta = -2 \), then \( \alpha = 8 \).
If \( \beta = -4 \), then \( \alpha = 4 \).
4. Magnitude \( |\vec{a}|^2 = 1^2 + 5^2 + \alpha^2 = 26 + \alpha^2 \).
Value 1: \( 26 + (8)^2 = 26 + 64 = 90 \).
Value 2: \( 26 + (4)^2 = 26 + 16 = 42 \).
The greatest value is 90.
Step 3: Final Answer:
The greatest value of \( |\vec{a}|^2 \) is 90.