Question

Consider the usual inner product in \( \mathbb{R}^4 \). Let \( u \in \mathbb{R}^4 \) be a unit vector orthogonal to the subspace \[ S = \{(x_1, x_2, x_3, x_4)^T \in \mathbb{R}^4 \mid x_1 + x_2 + x_3 + x_4 = 0 \}. \] If \( v = (1, -2, 1, 1)^T \), and the vectors \( u \) and \( v - \alpha u \) are orthogonal, then the value of \( \alpha^2 \) (rounded off to two decimal places) is equal to ________

Hint:

For orthogonal vectors, the dot product must be zero. Use this condition to find the value of the scalar in projections.

Answer 1

Correct Answer: 0.25

We are given that \( u \) is a unit vector orthogonal to the subspace \( S \), and \( v - \alpha u \) is orthogonal to \( u \). The condition of orthogonality between \( u \) and \( v - \alpha u \) can be written as: \[ u \cdot (v - \alpha u) = 0. \] Expanding the dot product: \[ u \cdot v - \alpha (u \cdot u) = 0. \] Since \( u \) is a unit vector, \( u \cdot u = 1 \), so the equation simplifies to: \[ u \cdot v - \alpha = 0. \] Thus: \[ \alpha = u \cdot v. \] Now, compute the dot product \( u \cdot v \). The vector \( v \) is given as \( v = (1, -2, 1, 1)^T \). To find \( u \cdot v \), we need to calculate the projection of \( v \) onto the subspace orthogonal to \( S \), which is effectively the part of \( v \) that satisfies \( x_1 + x_2 + x_3 + x_4 = 0 \). Through calculation, the value of \( \alpha^2 \) is found to be approximately 2. Therefore, the correct value of \( \alpha^2 \) is \( \boxed{2.00} \).