Equations of planes parallel to the plane x-2 | vedclass

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(D) The equation of a plane parallel to $x-2y+2z+4=0$ is of the form $x-2y+2z+k=0$.The distance $d$ from a point $(x_1, y_1, z_1)$ to the plane $Ax+By

Vedclass · Accepted Answer

$x-2y+2z=0, x-2y+2z-6=0$

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

Equations of planes parallel to the plane $x-2y+2z+4=0$ which are at a distance of $1$ unit from the point $(1, 2, 3)$ are $.....$

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