If a plane P passes through the points (1,0,0 | vedclass

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(B) Let the equation of the plane $P$ be $a(x-1) + by + cz = 0$, which simplifies to $ax + by + cz = a$. Since it passes through $(0,1,0)$, we have $b

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$1, 1, \sqrt{2}$

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(B) Let the equation of the plane $P$ be $a(x-1) + by + cz = 0$, which simplifies to $ax + by + cz = a$. Since it passes through $(0,1,0)$, we have $b = a$. Thus, the equation is $ax + ay + cz = a$, or $x + y + \frac{c}{a}z = 1$. Let $k = \frac{c}{a}$. The normal vector is $\vec{n_1} = (1, 1, k)$.The normal to the plane $x + y = 3$ is $\vec{n_2} = (1, 1, 0)$.The angle between the planes is $\frac{\pi}{4}$, so $\cos \frac{\pi}{4} = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$.$\frac{1}{\sqrt{2}} = \frac{|1+1+0|}{\sqrt{1^2+1^2+k^2} \sqrt{1^2+1^2+0^2}} = \frac{2}{\sqrt{2+k^2} \sqrt{2}}$.Squaring both sides: $\frac{1}{2} = \frac{4}{2(2+k^2)} \Rightarrow 2+k^2 = 4 \Rightarrow k^2 = 2 \Rightarrow k = \sqrt{2}$.Thus, the direction ratios are $(1, 1, \sqrt{2})$.

If a plane $P$ passes through the points $(1,0,0)$ and $(0,1,0)$ and makes an angle $\frac{\pi}{4}$ with the plane $x+y=3$, then the direction ratios of a normal to that plane $P$ are

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