Question

Consider the orthonormal set \[ v_1 = \begin{bmatrix} \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \end{bmatrix}, v_2 = \begin{bmatrix} \frac{1}{\sqrt{6}} \\ \frac{2}{\sqrt{6}} \\ \frac{1}{\sqrt{6}} \end{bmatrix}, v_3 = \begin{bmatrix} \frac{1}{\sqrt{2}} \\ 0 \\ \frac{1}{\sqrt{2}} \end{bmatrix} \] with respect to the standard inner product on \( \mathbb{R}^3 \). If \( u = \begin{bmatrix} a \\ b \\ c \end{bmatrix} \) is the vector such that inner products of \( u \) with \( v_1, v_2 \) and \( v_3 \) are 1, 2 and 3, respectively, then \( a^2 + b^2 + c^2 \) (in integer) equals ................

Hint:

- The inner product in \( \mathbb{R}^3 \) is computed as the sum of the product of corresponding components.
- The system of equations from the inner product conditions can be solved using standard linear algebra techniques.

Answer 1

We are given that the vector \( u = \begin{bmatrix} a
b
c \end{bmatrix} \) has inner products with \( v_1, v_2 \) and \( v_3 \) as 1, 2, and 3, respectively. The inner product condition translates into the following system of equations: \[ \langle u, v_1 \rangle = a . \frac{1}{\sqrt{3}} + b . \frac{1}{\sqrt{3}} + c . \frac{1}{\sqrt{3}} = 1 \] \[ \langle u, v_2 \rangle = a . \frac{1}{\sqrt{6}} + b . \frac{2}{\sqrt{6}} + c . \frac{1}{\sqrt{6}} = 2 \] \[ \langle u, v_3 \rangle = a . \frac{1}{\sqrt{2}} + b . 0 + c . \frac{1}{\sqrt{2}} = 3 \] Solving this system of linear equations will give us the values of \( a \), \( b \), and \( c \). After solving, we find that: \[ a^2 + b^2 + c^2 = 14 \] Final Answer: The value of \( a^2 + b^2 + c^2 \) is \( \boxed{14} \).