Let vec{p} = 2hat{i} + 3hat{j} + ahat{k},vec{ | vedclass

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(A) Since $\vec{p}, \vec{q}, \vec{r}$ are coplanar,their scalar triple product is zero: $\left[ \vec{p} \, \vec{q} \, \vec{r} \right] = 0$. This is gi

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$(1, 3)$ or $(13, 9)$

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(A) Since $\vec{p}, \vec{q}, \vec{r}$ are coplanar,their scalar triple product is zero: $\left[ \vec{p} \, \vec{q} \, \vec{r} ight] = 0$. This is given by the determinant: $\begin{vmatrix} 2 & 3 & a \ b & 5 & -1 \ 1 & 1 & 3 \end{vmatrix} = 0$. Expanding the determinant: $2(15 - (-1)) - 3(3b - (-1)) + a(b - 5) = 0$. $2(16) - 3(3b + 1) + ab - 5a = 0$. $32 - 9b - 3 + ab - 5a = 0 \Rightarrow ab - 5a - 9b + 29 = 0$. Given $\vec{p} \cdot \vec{q} = 20$: $(2\hat{i} + 3\hat{j} + a\hat{k}) \cdot (b\hat{i} + 5\hat{j} - \hat{k}) = 20$. $2b + 15 - a = 20 \Rightarrow 2b - a = 5 \Rightarrow a = 2b - 5$. Substitute $a = 2b - 5$ into the first equation: $(2b - 5)b - 5(2b - 5) - 9b + 29 = 0$. $2b^2 - 5b - 10b + 25 - 9b + 29 = 0$. $2b^2 - 24b + 54 = 0 \Rightarrow b^2 - 12b + 27 = 0$. $(b - 3)(b - 9) = 0$. If $b = 3$,then $a = 2(3) - 5 = 1$. If $b = 9$,then $a = 2(9) - 5 = 13$. Thus,the ordered pairs are $(1, 3)$ or $(13, 9)$.

Let $\vec{p} = 2\hat{i} + 3\hat{j} + a\hat{k}$,$\vec{q} = b\hat{i} + 5\hat{j} - \hat{k}$,and $\vec{r} = \hat{i} + \hat{j} + 3\hat{k}$. If $\vec{p}, \vec{q}, \vec{r}$ are coplanar and $\vec{p} \cdot \vec{q} = 20$,then the ordered pair $(a, b)$ is:

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