Question

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

• A. $2 x+3 y-4 z=3$
• B. $2 x+3 y-4 z=1 \pm 2 \sqrt{29}$
• C. $2 x+3 y-4 z=-1 \pm 2 \sqrt{29}$
• D. $2 x+3 y-4 z=-3$

Answer 1

Answer: (B)

$ r=x \hat{i}+y \hat{j}+z \hat{k}$
$ r \cdot(2 \hat{i}+3 \hat{j}-4 \hat{k})=1$
$ (x \hat{i}+y \hat{j}+z \hat{k})(2 \hat{i}+3 \hat{j}-4 \hat{k})=1 $
$2 x+3 y-4 z=1 \ldots \text { (i) }$
$\therefore \text { Equation of plane parallel to Eq. (i), } $
$ 2 x+3 y-4 z=k $
$\frac{\left|d_1-d_2\right|}{\sqrt{a^2+b^2+c^2}}=2 $
$ \frac{|k-1|}{\sqrt{4+9+16}}=2 $
$ k-1= \pm 2 \sqrt{29}$
$ k=1 \pm 2 \sqrt{29}$
Hence, equation of required plane
$2 x+3 y-4 z=1 \pm 2 \sqrt{29}$