# Find the Distance of the Point (2, 12, 5) from the Point of Intersection of the Line - Mathematics

Find the distance of the point (2, 12, 5) from the point of intersection of the line

vecr=2hati-4hat+2hatk+lambda(3hati+4hatj+2hatk)

#### Solution

Let the point of intersection of the line and the plane be(x1,y1,z1)

Equation of line is (x−2)/3=(y+4)/4=(z−2)/2 Point (x1,y1,z1) will satisfy the equation of the line

(x_1−2)/3=(y_1+4)/4=(z_1−2)/k

⇒x_1=3k+2; y_1=4k−4; z_1=2k+2                       ..........(1)

Equation of plane is x-2y+z=0.

Point (x1,y1,z​1) will also satisfy the equation of plane.

⇒x_1−2y_1+z_1=0

Substituting equation (1) in equation (2), we get:

3k+22(4k4)+2k+2=0

k=4

Hence,

x_1=3(4)+2;

 y_1=4(4)−4;

 z_1=2(4)+2

 x_1=14;

y_1=12;

z_1=10

Thus, the point of intersection is (14,12,10).

Distance between the point ​(14,12,10) and ​(2,12,5)=sqrt((x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2)

=sqrt((14-2)^2+(12-12)^2+(10-5)^2)

=sqrt(144+0+25)

=sqrt169

=13 units

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

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