Distance between the two planes: 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is A. 2 units B. 4 units C. 8 units D. 2/√29 units

Home » Distance between the two planes: 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is A. 2 units B. 4 units C. 8 units D. 2/√29 units

Distance between the two planes: 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is
A. 2 units
B. 4 units
C. 8 units
D. 2/√29 units

Distance between the two planes: 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is
A. 2 units
B. 4 units
C. 8 units
D. 2/√29 units

Solution:

It is known to us that the distance between two parallel planes $A x+B y+C z=d_{1}$ and $A x+B y+C z=d_{2}$ is given as
$\left|\frac{\mathrm{d}_{1}-\mathrm{d}_{2}}{\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}+\mathrm{C}^{2}}}\right|$
Given:
First Plane:
$2 x+3 y+4 z=4$
Let’s compare with $A x+B y+C z=d_{1}$ , we obtain
$A=2, B=3, C=4, d_{1}=4$
Second Plane:
$4 x+6 y+8 z=12$ [On dividing the eq. by 2 $]$
We obtain,
$2 x+3 y+4 z=6$
Now comparing with $A x+B y+C z=d_{1}$ , we obtain
$A=2, B=3, C=4, d_{2}=6$
Therefore,
The distance between two planes is given as
$\begin{array}{l} =\left|\frac{4-6}{\sqrt{2^{2}+3^{2}+4^{2}}}\right| \\ =\left|\frac{-2}{\sqrt{4+9+16}}\right| \\ =2 / \sqrt{29} \end{array}$
As a result, option (D) is the correct option.