# Find the distance of the point (−1, −5, −10) from the point of intersection of the line r=2i-j+2k+λ(3i+4j+2k) and the plane r (i-j+k)=5 - Mathematics

Find the distance of the point (−1, −5, −10) from the point of intersection of the line vecr=2hati-hatj+2hatk+lambda(3hati+4hatj+2hatk)  and the plane vec r (hati-hatj+hatk)=5

#### Solution

Let us suppose that the given line and plane intersect at the point P(x,y,z).

∴ The position vector of P is vecr=xhati+yhatj+zhatk

Thus, the given equations of the line and the plane can be rewritten as

xhati+yhatj+zhatk=(2+3λ)hati−(1+4λ)hayj+(2+2λ)hatk and (xhati+yhatj+zhatk).(hati−hatj+hatk) = 5, respectively.

On simplifying xhati+yhatj+zhatk=(2+3λ)hati−(1+4λ)hatj+(2+2λ)hatk and (xhati+yhatj+zhatk).(hati−hatj+hatk) = 5 , we get:

x=(2+3λ)y =(1+4λ)z = (2+2λ)
Also, xy+z = 5

On putting the values of x, y and z in the equation xy+z = 5, we get:

2+3λ+1+4λ+2+2λ = 59λ+5 = 5λ = 0

x=2, y =1 and z =2

Hence, the distance between the points (−1, −5, −10) and (2, −1, 2) is

sqrt((2+1)^2+(−1+5)^2+(2+10)^2)=sqrt(9+16+144)=sqrt169=13 units

Concept: Three - Dimensional Geometry Examples and Solutions
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2013-2014 (March) Delhi Set 1

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