Compute the length of a vector satisfying given relations

Given vectors $A,B \in \mathbb{R}^n$ such that $\lVert A \rVert = 6$ and such that for every pair $x,y \in \mathbb{R}$ the two vectors

$xA + yB \qquad \text{and} \qquad 4yA - 9xB$

are orthogonal. Compute the length of the vectors $B$ and $2A + 3B$ .

Let $A = (a_1, \ldots, a_n)$ and $B = (b_1, \ldots, b_n)$ . Then, since $xA + yB$ and $4yA - 9xB$ are orthogonal we know that for every choice of $x, y\in \mathbb{R}$ we have

$\begin{align*} (xA + yB) \cdot (4yA - 9xB) &= (xa_1 + yb_1, \ldots, xa_n + yb_n) \cdot (4ya_1 - 9xb_1, \ldots, 4ya_n - 9xb_n) \\ &= \sum_{i=1}^n \left( 4xya_i^2 + 4y^2 a_i b_i - 9x^2 a_i b_i - 9xy b_i^2 right) \\ &= 0. \end{align*}$

This must hold for all choices of $x$ and $y$ , so in particular, it must hold for $x = 2, y = 3$ . Substituting these values gives us the equation

$\begin{align*} && \sum_{i=1}^n \left( 24a_i^2 + 36a_i b_i - 36a_i b_i - 54 b_i^2 \right) &= 0\\[9pt] \implies && (24 \cdot 36) - 54 \cdot \lVert B \rVert^2 &= 0 \\[9pt] \implies && \lVert B \rVert = 4. \end{align*}$

Next, we have

$\lVert 2A+3B \rVert = \left( \sum_{i=1}^n (2A+3B) \cdot (2A+3B) \right)^{\frac{1}{2}} = \left( \sum_{i=1}^n (4a_i^2 + 12a_i b_i + 9b_i^2) \right)^{\frac{1}{2}}.$

Letting $x = y = 1$ in our orthogonality equation we have

$\begin{align*} && \sum_{i=1}^n (4a_i^2 - 5a_i b_i - 9b_i^2) &= 0 \\[9pt] \implies && 4 \cdot 36 - 9 \cdot 16 - 5 \sum_{i=1}^n a_i b_i &= 0 \\[9pt] \implies && \sum_{i=1}^n a_i b_i &= 0. \end{align*}$

Thus,

$\begin{align*} \lVert 2A+3B \rVert &= \left( \sum_{i=1}^n (4a_i^2 + 9b_i^2) \right)^{\frac{1}{2}} \\[9pt] &= \left( 4 \cdot 36 + 9 \cdot 16 \right)^{\frac{1}{2}} \\[9pt] &= \sqrt{288} \\ &= 12 \sqrt{2}. \end{align*}$

Stumbling Robot

Compute the length of a vector satisfying given relations

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