Q51 / 110 · VITEEE 2024 · Mathematics
If the distance between the planes $8 x+12 y-14 z=2$ and $4 x+6 y-7 z=2$ can be expressed as $\frac{1}{\sqrt{N}}$, then the value of $\frac{N(N+1)}{2}$ is
- A4950
- B5050
- C5150
- Correct answerD5151
Solution
$P_1: 4 x+6 y-7 z=2$
$P_2: 8 x+12 y-14 z=2 \Rightarrow 4 x+6 y-7 z=1$
Distance between $P_1$ and $P_2$
$\begin{aligned}
& =\left|\frac{d_2-d_1}{\sqrt{a_1 a_2+b_1 b_2+c_1 c_2}}\right| \\
& =\left|\frac{2-1}{\sqrt{4 \times 4+6 \times 6+(-7) \times(-7)}}\right| \\
& =\frac{1}{\sqrt{16+36+49}}=\frac{1}{\sqrt{101}}
\end{aligned}$
By comparing with $\frac{1}{\sqrt{N}}, N=101$
Therefore, $\frac{N(N+1)}{2}=\frac{101 \times 102}{2}=5151$
Hence, the correct answer is option: 'D'.