Question

The line $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$is parallel to the plane

• A. $3x+4y+5z=7$
• B. $2x+3y+4z=0$
• C. $x+y-z=2$
• D. $2x+y-2z=0$

Answer 1

The correct option is D

$2x+y-2z=0$

Explanation for the correct option:

Given: line $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$

let $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}=\lambda$

$3\hat{i}+4\hat{j}+5\hat{k}=\overrightarrow{a}$ will be perpendicular to normal vector of a plane.

Checking option D

So,

$$\begin{gathered} =3\left(2\right)+4\left(1\right)+5(-2) \\ =6+4-10 \\ =0 \end{gathered}$$

Hence, $2x+y-2z=0$ is parallel to $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$.

Explanation for the in-correct options:

Option A $3x+4y+5z=7$

$3\hat{i}+4\hat{j}+5\hat{k}=\overrightarrow{a}$ and normal vecotor to the plane is $3\hat{i}+4\hat{j}+5\hat{k}=\overrightarrow{n}$

Both vector are parallel so option (A) is incorrect.

Option(B) $2x+3y+4z=0$

$2\hat{i}+3\hat{j}+4\hat{k}=\overrightarrow{n}$ and $3\hat{i}+4\hat{j}+5\hat{k}=\overrightarrow{a}$

$\overrightarrow{a}.\overrightarrow{n}=6+12+20\ne 0$

So, option B is also incorrect.

Option(C) $x+y-z=2$

$\hat{i}+\hat{j}-\hat{k}=\overrightarrow{n}$ and $3\hat{i}+4\hat{j}+5\hat{k}=\overrightarrow{a}$

$\overrightarrow{a}.\overrightarrow{n}=3+4-5\ne 0$

So, option C is also incorrect.

Hence,option D is correct.