The equations of planes parallel to the plane | vedclass

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(A) The equation of any plane parallel to $x+2y+2z+8=0$ is of the form $x+2y+2z+\lambda=0$.Given that the distance of this plane from the point $(1,1,

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$x+2y+2z-13=0$ or $x+2y+2z-1=0$

Vedclass · Answer

(A) The equation of any plane parallel to $x+2y+2z+8=0$ is of the form $x+2y+2z+\lambda=0$.Given that the distance of this plane from the point $(1,1,2)$ is $2$ units.The distance $d$ of a point $(x_1, y_1, z_1)$ from the plane $Ax+By+Cz+D=0$ is given by $d = \frac{|Ax_1+By_1+Cz_1+D|}{\sqrt{A^2+B^2+C^2}}$.Substituting the values,we get:$2 = \frac{|1(1)+2(1)+2(2)+\lambda|}{\sqrt{1^2+2^2+2^2}}$$2 = \frac{|1+2+4+\lambda|}{\sqrt{1+4+4}}$$2 = \frac{|7+\lambda|}{\sqrt{9}}$$2 = \frac{|7+\lambda|}{3}$$|7+\lambda| = 6$This implies $7+\lambda = 6$ or $7+\lambda = -6$.Case $1$: $7+\lambda = 6 \Rightarrow \lambda = -1$.Case $2$: $7+\lambda = -6 \Rightarrow \lambda = -13$.Thus,the equations of the planes are $x+2y+2z-1=0$ and $x+2y+2z-13=0$.

The equations of planes parallel to the plane $x+2y+2z+8=0$,which are at a distance of $2$ units from the point $(1,1,2)$ are

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