QUESTION

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can someone please help
You are given the cross product \[ \left(\begin{array}{l} 3 \\ 1 \\ 4 \end{array}\right) \times\left(\begin{array}{l} 4 \\ 0 \\ 2 \end{array}\right)=\left(\begin{array}{c} 2 \\ 10 \\ -4 \end{array}\right) \] Let's observe that \[ \begin{aligned} \left(\begin{array}{l} 3 \\ 1 \\ 4 \end{array}\right) \cdot\left(\begin{array}{c} 2 \\ 10 \\ -4 \end{array}\right)=\text { Number } \\ \text { and }\left(\begin{array}{l} 4 \\ 0 \\ 2 \end{array}\right) \cdot\left(\begin{array}{c} 2 \\ 10 \\ -4 \end{array}\right)=\text { Number } \end{aligned} \] This means that $\left(\begin{array}{c}2 \\ 10 \\ -4\end{array}\right)$ is perpendicular $\quad$ to both $\left(\begin{array}{l}3 \\ 1 \\ 4\end{array}\right)$ and $\left(\begin{array}{l}4 \\ 0 \\ 2\end{array}\right)$, the vectors in the given cross product.
i) So, using this idea, a (non-zero) vector perpendicular to both $\left(\begin{array}{l}0 \\ 3 \\ 5\end{array}\right)$ and $\left(\begin{array}{l}-2 \\ -5 \\ -1\end{array}\right)$ is ii) Likewise, a (non-zero) vector $\mathbf{v}$ satisfying both $\mathbf{v} \cdot\left(\begin{array}{l}1 \\ 3 \\ 1\end{array}\right)=0$ and $\mathbf{v} \cdot\left(\begin{array}{c}1 \\ 2 \\ -2\end{array}\right)=0$ is 因国。

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