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

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(A) Let the equation of the plane be $a(x - 0) + b(y + 1) + c(z - 0) = 0$,which simplifies to $ax + by + cz + b = 0$.Since the plane passes through $(

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

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(A) Let the equation of the plane be $a(x - 0) + b(y + 1) + c(z - 0) = 0$,which simplifies to $ax + by + cz + b = 0$.Since the plane passes through $(0, 0, 1)$,we have $a(0) + b(0) + c(1) + b = 0$,so $c = -b$.The equation becomes $ax + by - bz + b = 0$,or $ax + b(y - z + 1) = 0$.The normal vector to this plane is $\vec{n_1} = (a, b, -b)$.The normal vector to the plane $y - z + 5 = 0$ is $\vec{n_2} = (0, 1, -1)$.The angle between the planes is $\frac{\pi}{4}$,so $\cos(\frac{\pi}{4}) = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$.$\frac{1}{\sqrt{2}} = \frac{|0 + b + b|}{\sqrt{a^2 + 2b^2} \sqrt{2}} = \frac{|2b|}{\sqrt{2(a^2 + 2b^2)}}$.Squaring both sides: $\frac{1}{2} = \frac{4b^2}{2(a^2 + 2b^2)} \implies a^2 + 2b^2 = 4b^2 \implies a^2 = 2b^2 \implies a = \pm \sqrt{2}b$.Case $1$: $a = \sqrt{2}b$. The equation is $\sqrt{2}bx + by - bz + b = 0 \implies \sqrt{2}x + y - z + 1 = 0$.Checking point $(\sqrt{2}, -1, 4)$: $\sqrt{2}(\sqrt{2}) + (-1) - 4 + 1 = 2 - 1 - 4 + 1 = -2 
eq 0$.Checking point $(-\sqrt{2}, 1, -4)$: $\sqrt{2}(-\sqrt{2}) + 1 - (-4) + 1 = -2 + 1 + 4 + 1 = 4 
eq 0$.Case $2$: $a = -\sqrt{2}b$. The equation is $-\sqrt{2}bx + by - bz + b = 0 \implies -\sqrt{2}x + y - z + 1 = 0$.Checking point $(\sqrt{2}, -1, 4)$: $-\sqrt{2}(\sqrt{2}) - 1 - 4 + 1 = -2 - 1 - 4 + 1 = -6 
eq 0$.Checking point $(-\sqrt{2}, 1, -4)$: $-\sqrt{2}(-\sqrt{2}) + 1 - (-4) + 1 = 2 + 1 + 4 + 1 = 8 
eq 0$.Re-evaluating: The plane is $ax + by - bz + b = 0$. For point $(\sqrt{2}, 1, 4)$,$\sqrt{2}a + b - 4b + b = 0 \implies \sqrt{2}a = 2b \implies a = \sqrt{2}b$. This matches Case $1$.

$A$ 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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