Question

The projection of line $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$ on the plane 2x + 3y + 4z = 0 is

• A. $\frac{x}{11}=\frac{y}{2}=\frac{z}{-7}$
• B. $\frac{x}{2}=\frac{y}{11}=\frac{z}{7}$
• C. $\frac{x}{7}=\frac{y}{2}=\frac{-z}{7}$
• D. None of these

Answer 1

The correct option is D None of these

We have,

The given line is $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}......\left(1\right)$

And the plane is $2x+3y+4z=0......\left(2\right)$

From equation (1) to,

$\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}=k$

So, the points are

$(k,2k+1,3k+2)$

Now, these points also lie on the given plane

$2x+3y+4z=0$

$\Rightarrow 2k+3(2k+1)+4(3k+2)=0$

$\Rightarrow 2k+6k+3+12k+8=0$

$\Rightarrow 20k=-11$

$\Rightarrow k=\frac{-11}{20}$

So, the points $(-\frac{11}{20},\frac{-1}{20},\frac{-13}{20})$

Now, the projection on line is

$\frac{x+\frac{11}{20}}{2}=\frac{y+\frac{1}{20}}{3}=\frac{z+\frac{13}{20}}{4}$

$\Rightarrow \frac{20x+11}{40}=\frac{20y+1}{60}=\frac{20z+13}{80}$

Hence, this is the answer.