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The two vectors \({\bf{u}}{\rm{ and }}{\bf{v}}\) are Orthogonal if:

\(\begin{aligned}{l}{\left\| {{\bf{u}} + {\bf{v}}} \right\|^2} = {\left\| {\bf{u}} \right\|^2} + {\left\| {\bf{v}} \right\|^2}\\{\rm{and}}\\{\bf{u}} \cdot {\bf{v}} = 0\end{aligned}\).

The given vector is, \({\bf{u}} = \left( {\begin{aligned}{*{20}{c}}5\\{ - 6}\\7\end{aligned}} \right)\) and \(W = \left\{ {x \in {\mathbb{R}^3}|{\bf{u}} \cdot {\bf{x}} = 0} \right\}\).

Since is a null space of the \(1 \times 3\) matrix \({{\bf{u}}^T}\).

Therefore, Theorem 2 can be used to verify that \(W\) is a subspace of \({\mathbb{R}^3}\), which is possible only, if \({\bf{u}} \cdot {\bf{x}} = 0\) or \({{\bf{u}}^T} \cdot {\bf{x}} = 0\), this shows that \(W\) is a null-pace of \({{\bf{u}}^T}\). Hence \(W\) is a subspace of \({\mathbb{R}^3}\).

As \(W\) has all the vectors which are perpendicular to \({\bf{u}}\). So find \({\bf{u}} \cdot {\bf{x}} = 0\) by letting \({\bf{x}} = \left( {\begin{aligned}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{aligned}} \right)\).

\(\begin{aligned}{c}\left( {\begin{aligned}{*{20}{c}}5\\{ - 6}\\7\end{aligned}} \right) \cdot \left( {\begin{aligned}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{aligned}} \right) = 0\\5{x_1} - 6{x_2} + 7{x_3} = 0\end{aligned}\)

So geometrically, the subspace \(W\) is a plane passing through the origin.