Analyzing the Setup
To find the image of the point
P(1,3,5) with respect to the plane
4x−5y+2z=8, we first express the plane in the standard form:
4x−5y+2z−8=0
The line connecting the point P and its image P′(α,β,γ) must be perpendicular to the plane. This implies the direction ratios of the line are proportional to the normal vector of the plane, which is (4,−5,2).
The Master Equation
We utilize the mirror image formula to determine the coordinates of the reflected point:
ax−x1=by−y1=cz−z1=−2a2+b2+c2ax1+by1+cz1+d
Here, (x1,y1,z1)=(1,3,5) and (a,b,c)=(4,−5,2). The constant d is −8.
Calculating the Displacement Ratio
First, we evaluate the numerator of the displacement term by substituting the point into the plane equation:
4(1)−5(3)+2(5)−8=4−15+10−8=−9
Next, we calculate the denominator, which is the sum of the squares of the normal vector components:
42+(−5)2+22=16+25+4=45
We define the ratio
k as follows:
k=−2×(45−9)=−2×(−51)=52
Determining the Coordinates
Using the ratio k=52, we solve for the coordinates (α,β,γ) individually:
For
α:
4α−1=52⇒α=1+58=513
For
β:
−5β−3=52⇒β=3−2=1
For
γ:
2γ−5=52⇒γ=5+54=529
Final Calculation
The sum of the coordinates is:
α+β+γ=513+1+529=513+5+29=547
The problem asks for the value of
5(α+β+γ):
5×(547)=47
The final result is 47.