Question

The line $\frac{x-4}{4}=\frac{y-12}{12}=\frac{z-8}{8}$, which of the following is not true

• A. lies in the $x-y+Z=0$
• B. IS same as the Ime $\frac{x}{4}=\frac{y}{12}=\frac{z}{8}$
• C. is parallel to the plane $x-y+z-6=0$
• D. None of these

Answer 1

The correct option is D None of these

Given $L_1:\frac{x-4}{4}=\frac{y-12}{12}=\frac{z-8}{8}$\\ $D{R}^{\prime }s(4,12,8)=\overrightarrow{r}$

$A:Plane-P_1:x-y+z=0$

DR's of normal $(1,-1,1)=\overrightarrow{n}$

$\overrightarrow{r}\cdot \overrightarrow{n}=4-12+8=0$

Plane and line are parallel

So plane contains the line

Option B

$L_2:\frac{x}{4}=\frac{y}{12}=\frac{z}{8}$

If $L_1,L_2$ are same they should have same DR's and point of $L_1$ should lie on $L_2$ and point of $L_2$ should lie on $L_1$

$P_{1}in{L}_2:\frac{0-4}{4}=\frac{0-12}{12}=\frac{8}{8}$

Hence $P_{ 1 }$liesin${ L }_{ 2 }$

$P_{2}in{L}_1:\frac{0-4}{4}=\frac{0-12}{12}=\frac{8}{8}$

Hence $P_{ 2 }$liesin${ L }_{ 1 }$

So ${ L }_{ 1 }$and${ L }_{ 2 }$ are same.

Option C

Plane ${ P }_{ 1 }$and${ L }_{ 1 }$ are parallel

$x-y+z=6$

If normal is perpendicular to line

$\left(4\right)1-12+8=0$

Hence plane $x-y+z-6=0$ is parallel to $L_1$