Q. 20

Find the distance of the point (–1, –5, –10), from the point of intersection of the line and the plane [CBSE 2011]

Answer :

Given, the equation of the line is –



The cartesian form of the equation is –



x = 3λ + 2 …(1)


y = 4λ – 1 …(2)


and z = 2λ + 2 …(3)


Given equation of plane is –



The cartesian form of above equation is –


x – y + z = 5 …(4)


At the point of intersection of line and the plane-


We have –


3λ +2 – (4λ – 1) + (2λ + 2) = 5


λ + 5 = 5


λ = 0


Hence, x = 2 ; y = -1 and z = 2


position vector of the intersection point is or coordinate is (2,-1,2)


By distance formula, we know that distance between two points (x1,y1,z1) and (x2,y2,z2) is given by –



distance between (-1, -5, -10) and (2,-1, 2) is given by –



D = = 13 units.


Hence, a distance of the point (–1, –5, –10), from the point of intersection of the line and the plane is 13 units

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