Let a unit vector hat{OP} make angles alpha, | vedclass

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Question

(A) The equation of the plane passing through points $A(1, 2, 3)$,$B(2, 3, 4)$,and $C(1, 5, 7)$ is given by the determinant:$\left|\begin{array}{ccc}

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$\alpha \in (\frac{\pi}{2}, \pi)$ and $\gamma \in (\frac{\pi}{2}, \pi)$

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(A) The equation of the plane passing through points $A(1, 2, 3)$,$B(2, 3, 4)$,and $C(1, 5, 7)$ is given by the determinant:$\left|\begin{array}{ccc} x-1 & y-2 & z-3 \ 2-1 & 3-2 & 4-3 \ 1-1 & 5-2 & 7-3 \end{array}ight| = 0$$\Rightarrow \left|\begin{array}{ccc} x-1 & y-2 & z-3 \ 1 & 1 & 1 \ 0 & 3 & 4 \end{array}ight| = 0$Expanding along the first row: $(x-1)(4-3) - (y-2)(4-0) + (z-3)(3-0) = 0$$\Rightarrow (x-1) - 4(y-2) + 3(z-3) = 0$$\Rightarrow x - 4y + 3z = 2$The normal vector to the plane is $\vec{n} = \langle 1, -4, 3 angle$. The unit normal vector $\hat{n}$ is $\pm \frac{\langle 1, -4, 3 angle}{\sqrt{1^2 + (-4)^2 + 3^2}} = \pm \frac{\langle 1, -4, 3 angle}{\sqrt{26}}$.Since $\hat{OP}$ is a unit vector perpendicular to the plane,$\hat{OP} = \pm \langle \frac{1}{\sqrt{26}}, \frac{-4}{\sqrt{26}}, \frac{3}{\sqrt{26}} angle$. The direction cosines are $\cos \alpha, \cos \beta, \cos \gamma$. Given $\beta \in (0, \frac{\pi}{2})$,we must have $\cos \beta > 0$. Thus,$\cos \beta = \frac{-(-4)}{\sqrt{26}} = \frac{4}{\sqrt{26}}$.This implies $\hat{OP} = \langle -\frac{1}{\sqrt{26}}, \frac{4}{\sqrt{26}}, -\frac{3}{\sqrt{26}} angle$.Therefore,$\cos \alpha = -\frac{1}{\sqrt{26}} < 0 \Rightarrow \alpha \in (\frac{\pi}{2}, \pi)$ and $\cos \gamma = -\frac{3}{\sqrt{26}} < 0 \Rightarrow \gamma \in (\frac{\pi}{2}, \pi)$.

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