Q. 184.5( 19 Votes )
Find the distance of the point (–1, –5, –10) from the point of intersection of the line 
and the plane 
.
Answer :
Given -
The equation of line is
and the equation of the plane is
To find the intersection of line and plane, putting value of 
from equation of line into equation of plane, we get -
⇒ 
⇒ 
⇒ (2 + 3λ) × 1 + (-1 + 4λ) × (-1) + (2 + 2λ) × 1 = 5
⇒ 2 + 3λ + 1 - 4λ + 2 + 2λ = 5
⇒ λ = 0
So, the equation of line is
Let the point of intersection be (x,y,z)
So, 
∴ 
Hence, x = 2, y = -1, z = 2
Therefore, the point of intersection is (2, -1, 2).
Now, the distance between points (x1, y1, z1) and (x2, y2, z2) is given by -
units
Distance between the points A(2, -1, 2) and B(-1, -5, -10) is given by -
= 13 units
Rate this question :
Find the equation of the line passing through the point (–1, 3, –2) and perpendicular to the lines 
and 
Find the length of the perpendicular drawn from the origin to the plane 2x – 3y + 6z + 21 = 0.
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. Also find the image of P in this line.
Show that the lines 
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