The direction ratios of the normal to the pla | vedclass

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(D) 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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$\sqrt{13}, \sqrt{13}, 2\sqrt{6}$

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(D) 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 normal vector is $\vec{n_1} = (a, a, c)$.The normal to the plane $2x + 3y = 7$ is $\vec{n_2} = (2, 3, 0)$.The angle $	heta = 45^{\circ}$ between the planes is given by $\cos(45^{\circ}) = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$.$\frac{1}{\sqrt{2}} = \frac{|2a + 3a + 0|}{\sqrt{a^2 + a^2 + c^2} \sqrt{2^2 + 3^2}} = \frac{|5a|}{\sqrt{2a^2 + c^2} \sqrt{13}}$.Squaring both sides: $\frac{1}{2} = \frac{25a^2}{13(2a^2 + c^2)}$.$13(2a^2 + c^2) = 50a^2 \implies 26a^2 + 13c^2 = 50a^2 \implies 13c^2 = 24a^2 \implies c^2 = \frac{24}{13}a^2 \implies c = \pm a \sqrt{\frac{24}{13}} = \pm a \frac{2\sqrt{6}}{\sqrt{13}}$.Taking $a = \sqrt{13}$, we get $b = \sqrt{13}$ and $c = \pm 2\sqrt{6}$. Thus, the direction ratios are $\sqrt{13}, \sqrt{13}, 2\sqrt{6}$.

The direction ratios of the normal to the plane passing through $(1, 0, 0)$ and $(0, 1, 0)$ which makes an angle of $\pi/4$ with the plane $2x + 3y = 7$ are

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