A plane passing through the points (0, -1, 0) | vedclass

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(A) Let the equation of the plane be $ax + by + cz = d$. Since it passes through $(0, -1, 0)$,we have $-b = d$. Since it passes through $(0, 0, 1)$,we

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$(\sqrt{2}, 1, 4)$

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(A) Let the equation of the plane be $ax + by + cz = d$. Since it passes through $(0, -1, 0)$,we have $-b = d$. Since it passes through $(0, 0, 1)$,we have $c = d$. Let $d = 1$,then $b = -1$ and $c = 1$. The equation is $ax - y + z = 1$.The normal vector to this plane is $\vec{n_1} = (a, -1, 1)$ and the normal vector to the plane $y - z + 5 = 0$ is $\vec{n_2} = (0, 1, -1)$.The angle $	heta = \frac{\pi}{4}$ between the planes is given by $\cos 	heta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$.$\frac{1}{\sqrt{2}} = \frac{|(a)(0) + (-1)(1) + (1)(-1)|}{\sqrt{a^2 + (-1)^2 + 1^2} \sqrt{0^2 + 1^2 + (-1)^2}}$.$\frac{1}{\sqrt{2}} = \frac{|-2|}{\sqrt{a^2 + 2} \sqrt{2}}$.$\sqrt{a^2 + 2} = 2 \implies a^2 + 2 = 4 \implies a^2 = 2 \implies a = \pm \sqrt{2}$.Taking $a = \sqrt{2}$,the equation is $\sqrt{2}x - y + z = 1$. Checking point $(\sqrt{2}, 1, 4)$: $\sqrt{2}(\sqrt{2}) - 1 + 4 = 2 - 1 + 4 = 5 
eq 1$.Taking $a = -\sqrt{2}$,the equation is $-\sqrt{2}x - y + z = 1$. Checking point $(\sqrt{2}, -1, 4)$: $-\sqrt{2}(\sqrt{2}) - (-1) + 4 = -2 + 1 + 4 = 3 
eq 1$. Checking point $(-\sqrt{2}, 1, -4)$: $-\sqrt{2}(-\sqrt{2}) - 1 + (-4) = 2 - 1 - 4 = -3 
eq 1$. Checking point $(\sqrt{2}, 1, 4)$ in $-\sqrt{2}x - y + z = 1$: $-\sqrt{2}(\sqrt{2}) - 1 + 4 = -2 - 1 + 4 = 1$. Thus,the plane passes through $(\sqrt{2}, 1, 4)$.

$A$ plane passing through the points $(0, -1, 0)$ and $(0, 0, 1)$ and making an angle $\frac{\pi}{4}$ with the plane $y - z + 5 = 0$ also passes through the point

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