Vector Algebra
Magnitude and Inequality
Grade 12
Question:
<p>If \(a^2 + b^2 + c^2 = 1\) where \(a, b, c \in \mathbb{R}\), then the maximum value of \((4a - 3b)^2 + (5b - 4c)^2 + (3c - 5a)^2\) is</p>
<p>(a) 25</p>
<p>(b) 50</p>
<p>(c) 144</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Express the sum as a cross product magnitude and apply the Cauchy-Schwarz inequality via cross product.
Step 1: Let \(\vec{r_1} = a\hat{i} + b\hat{j} + c\hat{k}\) and \(\vec{r_2} = 3\hat{i} + 4\hat{j} + 5\hat{k}\) Step 2: Compute the cross product: \(\vec{r_1} \times \vec{r_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a & b & c \\ 3 & 4 & 5 \end{vmatrix} = \hat{i}(5b - 4c) + \hat{j}(3c - 5a) + \hat{k}(4a - 3b)\) Step 3: Using the inequality \(|\vec{r_1} \times \vec{r_2}|^2 \leq |\vec{r_1}|^2|\vec{r_2}|^2\): \((5b - 4c)^2 + (3c - 5a)^2 + (4a - 3b)^2 \leq 1 \cdot (9 + 16 + 25) = 50\) ∴ Answer is (b).
Correct Answer: B