The direction ratios of the normal to the pla | vedclass

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

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$1, 1, \sqrt{2}$

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

The direction ratios of the normal to the plane passing through the points $(1, 0, 0)$ and $(0, 1, 0)$ and making an angle of $\pi/4$ with the plane $x + y = 3$ are proportional to:

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