The direction ratios of the normal to the pla | vedclass

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Question

(B) The equation of a plane passing through $(1, 0, 0)$ and $(0, 1, 0)$ can be written in intercept form as $\frac{x}{1} + \frac{y}{1} + \frac{z}{c} =

Vedclass · Accepted Answer

$1, 1, \sqrt{2}$

Vedclass · Answer

(B) The equation of a plane passing through $(1, 0, 0)$ and $(0, 1, 0)$ can be written in intercept form as $\frac{x}{1} + \frac{y}{1} + \frac{z}{c} = 1$,where $c$ is the $z$-intercept.This simplifies to $x + y + \frac{z}{c} = 1$.The direction ratios of the normal to this plane are $(1, 1, \frac{1}{c})$.The given plane is $x + y = 3$,and its normal vector has direction ratios $(1, 1, 0)$.The angle $	heta$ between two planes with normals $\vec{n_1} = (a_1, b_1, c_1)$ and $\vec{n_2} = (a_2, b_2, c_2)$ is given by $\cos 	heta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$.Given $	heta = \frac{\pi}{4}$,we have $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$.Substituting the values: $\frac{1}{\sqrt{2}} = \frac{|1(1) + 1(1) + \frac{1}{c}(0)|}{\sqrt{1^2 + 1^2 + (\frac{1}{c})^2} \sqrt{1^2 + 1^2 + 0^2}}$.$\frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2 + \frac{1}{c^2}} \cdot \sqrt{2}}$.$\frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2} \sqrt{2 + \frac{1}{c^2}}}$.$1 = \frac{2}{\sqrt{2 + \frac{1}{c^2}}}$.$\sqrt{2 + \frac{1}{c^2}} = 2$.Squaring both sides: $2 + \frac{1}{c^2} = 4$.$\frac{1}{c^2} = 2 \Rightarrow c^2 = \frac{1}{2} \Rightarrow c = \frac{1}{\sqrt{2}}$.The direction ratios are $(1, 1, \frac{1}{c}) = (1, 1, \sqrt{2})$.

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

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