Prove or disapprove: ### Problem Statement**Given**: \( \mathbf{w} = \langle w_1, w_2, w_3 \rangle \)**Prove or disapprove**: \( \mathbf{w} \cdot \mathbf{w} = \|\mathbf{w}\|^2 \).**Task**: Explain all details.### ExplanationTo address the problem, we need to understand the meanings of both dot product and vector norm:1. **Dot Product** (\(\mathbf{w} \cdot \mathbf{w}\)): - The dot product of a vector with itself is calculated by multiplying each of its components by itself and then summing the results: \[ \mathbf{w} \cdot \mathbf{w} = w_1^2 + w_2^2 + w_3^2 \]2. **Norm of a Vector** (\(\|\mathbf{w}\|\)): - The norm (or magnitude) of a vector is calculated as the square root of the sum of the squares of its components: \[ \|\mathbf{w}\| = \sqrt{w_1^2 + w_2^2 + w_3^2} \]3. **Square of the Norm** (\(\|\mathbf{w}\|^2\)): - Taking the square of the vector norm cancels out the square root: \[ \|\mathbf{w}\|^2 = (\sqrt{w_1^2 + w_2^2 + w_3^2})^2 = w_1^2 + w_2^2 + w_3^2 \]From the above calculations, it is evident that:\[\mathbf{w} \cdot \mathbf{w} = \|\mathbf{w}\|^2\]### ConclusionThe statement \( \mathbf{w} \cdot \mathbf{w} = \|\mathbf{w}\|^2 \) is indeed true. The dot product of a vector with itself is equal to the square of its norm.

Name: Prove or disapprove: ### Problem Statement**Given**: \( \mathbf{w} = \
Uploaded: 2024-03-20T06:38:44.000Z
Channel: Bartleby
Description: Prove or disapprove: ### Problem Statement**Given**: \( \mathbf{w} = \langle w_1, w_2, w_3 \rangle \)**Prove or disapprove**: \( \mathbf{w} \cdot \mathbf{w} = \|\mathbf{w}\|^2 \).**Task**: Explain all details.### ExplanationTo address the problem, we