If $\bar{a}=3 \hat{i}+\hat{j}-\hat{k}, \bar{b}=2 \hat{i}-\hat{j}+23 \hat{k}$ and $\bar{c}=7…
If $\bar{a}=3 \hat{i}+\hat{j}-\hat{k}, \bar{b}=2 \hat{i}-\hat{j}+23 \hat{k}$ and $\bar{c}=7 \hat{i}-\hat{j}+23 \hat{k}$, then which of the following is valid.
$\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}$ are mutually perpendicular
$\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}$ are non-coplanar
$\overline{\mathrm{a}}$ and $\overline{\mathrm{b}}$ are collinear
$\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}$ are coplanar
Solution
We have $\bar{a}=3 \hat{i}+\hat{j}-\hat{k}, \bar{b}=2 \hat{i}-\hat{j}+23 \hat{k}$ and $\bar{c}=7 \hat{i}-\hat{j}+23 \hat{k}$ $\bar{a} \cdot \bar{b}=6-1-23 \neq 0$ and $\bar{b} \cdot \bar{c}=14+1+529 \neq 0$
Thus $\bar{a}, \bar{b}, \bar{c}$ are non mutually perpendicular.
Also for $\overline{\mathrm{a}}$ and $\overline{\mathrm{b}}, \frac{3}{2} \neq-1 \neq \frac{-1}{23}$.
Thus $\bar{a}$ and $\bar{b}$ are not collinear.
Now $\left|\begin{array}{ccc}3 & 1 & -1 \\ 2 & -1 & 23 \\ 7 & -1 & 23\end{array}\right|=3(-23+23)-(46-161)-(-2+7) \neq 0$.
Thus $\bar{a}, \bar{b}, \bar{c}$ are non coplanar.