Find the image of the point $(1,3,4)$ on the plane $2x-y+z+3=0$.

Question

Find the image of the point $(1,3,4)$ on the plane $2x-y+z+3=0$.

Let the image be $(a,b,c)$.

Equation of the line joining $(1,3,4)$ and $(a,b,c)$ is $(a-1,b-3,c-4)$ and it will be parallel to the plane given.

Now we have $\dfrac{a-1}{2}=\dfrac{b-3}{-1}=\dfrac{c-4}{1}=k\implies a=2k+1,b=-k+3,c=k+4$.

Since $(a,b,c)$ lies on the plane so we get $k=-1$.

Hence the point will be $(-1,4,3)$.

But the answer is not coming .Where am I wrong??

Answer 1

Let $P=a(1,3,4)$ then

$$2a-3a+4a+3=0\implies a=-1$$

thus $$P=(-1,-3,-4)$$