Question

Find the image of the point P(3,5,7) in the plane 2x+y+z=16.

Answer 1

Step 1:

The cartesian equation of the line is x1=y−12=z−23x1=y−12=z−23----(1)

The given point is (1,6,3)

To find the image of P(1,6,3) in the line draw a line PR ⊥⊥ to the line .

Let R be the image of P and Q is the mid point of PR



Let a,b,ca,b,c be the direction cosines of PR.Since PR is ⊥⊥to the line,apply the condition of perpendicularity.

a×1+b×2+c×3=0a×1+b×2+c×3=0-------(2)

a+2b+3c=0a+2b+3c=0

Step 2:

Now let x−1a=y−6b=z−3c=x−1a=y−6b=z−3c=kk(say)------(3)

Any point on the line (2) is (ak+1,bk+6,ck+3)(ak+1,bk+6,ck+3)

Let the point be Q

But Q also lies on line (1)

∴ak+11=bk+6−12=ck+3−23∴ak+11=bk+6−12=ck+3−23

⇒ak+11=bk+52=ck+13⇒ak+11=bk+52=ck+13

⇒1(ak+1)+2(bk+5)+3(k+1)1×1+2×2+3×3⇒1(ak+1)+2(bk+5)+3(k+1)1×1+2×2+3×3

⇒14+(a+2b+3c)k14⇒14+(a+2b+3c)k14=1=1

⇒ak=0,bk=−3⇒ak=0,bk=−3 and ck=2ck=2

Q=(0+1,−3+6,2+3)Q=(0+1,−3+6,2+3)

=(1,3,5)=(1,3,5)

Step 3:

Since QQ is the midpoint of PR

1+x′21+x′2=1=1 and 6+y′26+y′2=3=3 and 3+z′23+z′2=5=5

⇒x′=1,y′=0⇒x′=1,y′=0 and z′=7z′=7

Hence (1,0,7) is the image of P in line (1)