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# I need help with - Three Dimensional Geometry - JEE Main-6

Equation of plane parallel to $2x-y+2z-5=0$  at a distance of 1 unit from it is

• Option 1)

$2x-y-2z+8=0$

• Option 2)

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

• Option 3)

$2x-y+2z+8=0$

• Option 4)

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

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As we have learned

Distance between parallel planes (Cartesian form ) -

Distance between two planes $ax+by+cz+d= 0$ and

$ax+by+cz+d_{1}= 0$ is given by

$\frac{\left [ d_{1}-d\right ]}{\left | \sqrt{a^{2}+b^{2}+c^{2}} \right |}$

-

Let equation of plane is $2x-y+2z+k=0$ distance between them is $\frac{|k+5|}{3}=1\Rightarrow k+5=3,-3$

$\therefore k=-8,-2$

$\therefore$ Planes are $2x-y+2z-8=0$ and $2x-y+2z-2=0$

Option 1)

$2x-y-2z+8=0$

Option 2)

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

Option 3)

$2x-y+2z+8=0$

Option 4)

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

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