Find the coordinates of the point where the line through the points A(3, 4, 1) and B(5, 1, 6) crosses the XZ plane. Also find the angle which this line makes with the XZ plane.

#### Solution

The equation of the line through A(3, 4, 1) and B(5, 1, 6) is

`(x-3)/(5-3)=(y-4)/(1-4)=(z-1)/(6-1)`

or

`(x-3)/2=(y-4)/-3=(z-1)/5 .......................(1)`

Any point on (1) is given by (2*k* + 3, −3*k* + 4, 5*k* + 1).

We know that the coordinates of any point on the XZ plane are (*x*_{1}, 0, *z*_{1}).

If (2*k* + 3, −3*k* + 4, 5*k* + 1) lies on the XZ plane, then

If (2*k* + 3, −3*k* + 4, 5*k* + 1) lies on the XZ plane, then

−3*k* + 4 = 0

⇒k=43

Thus, the coordinates of the point where the line joining A and B crosses the XZ plane are

`(2xx4/3+3, 0, 5xx4/3+1)=(17/3, 0, 23/3)`

The vector equation of the XZ plane is `vecr.hatj=1`

The vector equation of the line AB is vecr=(3hati+4hatj+hatk)+lambda(2hati-3hatj+5hatk)

Let *θ *be the angle between the XZ plane and the line AB.

`therefore sin theta=(vecb vecn)/(|vecb||vecn|)=((2hati-3hatj+5hatk).hatj)/(sqrt(4+9+25)sqrt(0+1+0))`

`sin theta=(0-3+0)/sqrt(38)`

`=>theta=sin^-1 (-3/sqrt38)=-sin^-1 (3/sqrt38)`