Question

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

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

Answer 1

Answer: (B)

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