If the vector $\mathbf{v}$ is perpendicular to three non coplanar vectors, then prove that $\mathbf{v}=0$.
Solution:
Let $\mathbf{x} \in \mathbb{R}^3$ be an arbitrary vector. Let us denote the three non coplanar vectors as $\mathbf{a, b, c}$ respectively. Since they are not coplanar , this means that they are linearly independant and since they are three in total they will form a basis of $\mathbb{R}^3$. Hence the arbitrary vector $\mathbf{x}$ will be written as linear combination of these three vectors, that is there exist $\kappa, \lambda, \mu \in \mathbb{R}$ such that:
$$ \mathbf{x}= \kappa \mathbf{a} + \lambda \mathbf{b} + \mu \mathbf{c}$$
Taking inner product of $ \mathbf{v}$ and $\mathbf{x}$ and at the same time making use of the linearity of the inner product we get that the inner product of $\mathbf{x}$ and $\mathbf{v}$ is $0$, meaning that $\mathbf{v}$ is perpendicular to all vectors. More specifically for $\mathbf{x} = \mathbf{v}$ we get that $\mathbf{v \cdot v}=0$. This proves our claim .
Solution:
Let $\mathbf{x} \in \mathbb{R}^3$ be an arbitrary vector. Let us denote the three non coplanar vectors as $\mathbf{a, b, c}$ respectively. Since they are not coplanar , this means that they are linearly independant and since they are three in total they will form a basis of $\mathbb{R}^3$. Hence the arbitrary vector $\mathbf{x}$ will be written as linear combination of these three vectors, that is there exist $\kappa, \lambda, \mu \in \mathbb{R}$ such that:
$$ \mathbf{x}= \kappa \mathbf{a} + \lambda \mathbf{b} + \mu \mathbf{c}$$
Taking inner product of $ \mathbf{v}$ and $\mathbf{x}$ and at the same time making use of the linearity of the inner product we get that the inner product of $\mathbf{x}$ and $\mathbf{v}$ is $0$, meaning that $\mathbf{v}$ is perpendicular to all vectors. More specifically for $\mathbf{x} = \mathbf{v}$ we get that $\mathbf{v \cdot v}=0$. This proves our claim .
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