Question

Find the coordinates of the point where the line through the points A(3,4,1) and B(5,1,6)crosses the XY-plane.

Answer 1

Equation of a line passing through the points (x1,y1,z1)(x1,y1,z1) and (x2,y2,z2)(x2,y2,z2) is

x−x1x2−x1=y−y1y2−y1=z−z1z2−z1x−x1x2−x1=y−y1y2−y1=z−z1z2−z1

Equation of YZYZ plane is x=0x=0

Step 1:

We know that the equation of line passing through the points (x1,y1,z1)(x1,y1,z1) and (x2,y2,z2)(x2,y2,z2) is

x−x1x2−x1=y−y1y2−y1=z−z1z2−z1x−x1x2−x1=y−y1y2−y1=z−z1z2−z1

The line passing through the points (5,1,6)(5,1,6) and (3,4,1)(3,4,1) is given by,

Substituting for (x1,y1,z1)(x1,y1,z1) and (x2,y2,z2)(x2,y2,z2)

x−53−5=y−14−1=z−61−6x−53−5=y−14−1=z−61−6

(i.e) x−5−2=y−13=z−6−5x−5−2=y−13=z−6−5

Step 2:

Let this be equal to kk

x−5−2=y−13=z−6−5=x−5−2=y−13=z−6−5=kk

Therefore x=5−2kx=5−2k

y=3k+1y=3k+1

z=6−5kz=6−5k

Let the coordinates of this point be (5−2k,3k+1,6−5k)(5−2k,3k+1,6−5k)

Step 3:

The equation of YZYZ plane,x=0x=0

Since the line passes through the YZYZ plane.

5−2k=05−2k=0

⇒k=52⇒k=52

Step 4:

Now substituting for kk we get the coordinates as

(5−2.52,(5−2.52,3.523.52+1,6−5.52)+1,6−5.52)

On simplifying we get

(0,172(0,172,−132)

Answer 2

Given The Points Are A(3,4,1) B(5,1,6)
And Then Let Us Consider The Point C(a,b,0) Because Of Given That Crosses The XY-Plane
You Know That The General Equation Of Triple Variables
AX+BY+CZ+D=0 =>1
The Equation Passing Through Point A(3,4,1) Is
3A+4B+C+D=0 =>2
The Equation Passing Through Point B(5,1,6) Is
5A+B+6C+D=0 =>3
And Then The Equation Passing Through Point C(a,b,0) Is
aA+bB+0+D=0 => aA+bB+D=0 =>4
By Solving Equations 2,3 And 4
3A+4B+C+D=0
5A+B+6C+D=0
aA+bB+D=0 => aA+bB=-D
You Get The Required Point