The Cartesian equation of a plane parallel to | vedclass

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(B) The given plane is $\vec{r} \cdot(2 \hat{i}+3 \hat{j}-4 \hat{k})=1$. Converting this to Cartesian form, we get $2 x+3 y-4 z=1$, or $2 x+3 y-4 z-1=

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$2 x+3 y-4 z=1 \pm 2 \sqrt{29}$

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(B) The given plane is $\vec{r} \cdot(2 \hat{i}+3 \hat{j}-4 \hat{k})=1$. Converting this to Cartesian form, we get $2 x+3 y-4 z=1$, or $2 x+3 y-4 z-1=0$. Any plane parallel to this plane is of the form $2 x+3 y-4 z+\lambda=0$. The distance $d$ between two parallel planes $Ax+By+Cz+D_1=0$ and $Ax+By+Cz+D_2=0$ is given by $d = \frac{|D_1-D_2|}{\sqrt{A^2+B^2+C^2}}$. Here, $d=2$, $A=2$, $B=3$, $C=-4$, $D_1=-1$, and $D_2=\lambda$. So, $2 = \frac{|\lambda-(-1)|}{\sqrt{2^2+3^2+(-4)^2}} = \frac{|\lambda+1|}{\sqrt{4+9+16}} = \frac{|\lambda+1|}{\sqrt{29}}$. This implies $|\lambda+1| = 2 \sqrt{29}$, so $\lambda+1 = \pm 2 \sqrt{29}$, which means $\lambda = -1 \pm 2 \sqrt{29}$. Substituting $\lambda$ back into the equation $2 x+3 y-4 z+\lambda=0$, we get $2 x+3 y-4 z-1 \pm 2 \sqrt{29} = 0$, or $2 x+3 y-4 z = 1 \mp 2 \sqrt{29}$. Since the options provide $1 \pm 2 \sqrt{29}$, the correct equation is $2 x+3 y-4 z = 1 \pm 2 \sqrt{29}$.

The Cartesian equation of a plane parallel to the plane $\vec{r} \cdot(2 \hat{i}+3 \hat{j}-4 \hat{k})=1$ and at a distance of $2$ units from it is

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