Question

Find the reflection of the point (1,2,-1) in the plane 3x-5y+4z = 5

Answer 1

Answer:

33

Step-by-step explanation:

Answer 2

Therefore the reflection point is $(\frac{73}{25},\frac{-6}{5},\frac{39}{25})$

Step-by-step explanation:

Let $Q$ be the image of the point $P(1,2,-1)$ in the plane $3x-5y+4z=5$

Then $PQ$  is normal to the plane. So, the direction ratios of $PQ$ are proportional to $3,-5,4$.

Since $PQ$  passes through $P(1,2,-1)$ and has direction ratios proportional to $3,-5,4$, equation of $PQ$  is $x-13=y-2−5=z+14=r$ (say)

Let the coordinates of $Q$ be $(3r+1, -5r+2, 4r-1)$. Let $R$ be the mid-point of $PQ$.

Then,

$R=(3r + 1 + 12, -5r + 2 + 22, 4r-1-12) = (3r + 22, -5r + 42, 4r-22)$

Since $R$ lies in the plane $3x-5y + 4z = 5$,

$3 (3r + 22) -5 (-5r + 42) +4 (4r -22) = 5$

$9r + 6 + 25r -20+ 16r -8 = 10$

$50r -32 = 0$

$r=\frac{32}{50} =\frac{16}{25}$

Substituting the value of $r$ in the coordinates of $Q$, we get

$Q=(3r + 1, -5r + 2, 4r-1) = (3 (\frac{16}{25})+1, -5 (\frac{16}{25})+2, 4 (\frac{16}{25})-1) = (\frac{73}{25},\frac{-6}{5},\frac{39}{25})$

So the reflection point is $(\frac{73}{25},\frac{-6}{5},\frac{39}{25})$