Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12
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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

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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`

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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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