Find the vector equation of a line passing through the point with position vector 
and perpendicular to the plane 
+ 2 = 0.
Also, find the point of intersection of this line and the plane
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
= 
+ t 
, 
Sol. The required line is to pass through 
= 
and is ⊥ to the plane given by
+ 2 = 0 ...(1)
Here the vector 
= 
is normal to the plane (1)
⇒ The reqd. line is along the direction of this vector. Hence its equation is
= 
⇒ 
= 
+ t 
...(2)
Now line (2) meets the plane (1), when 
. 
= – 2
i.e. when 6(2 + 6t) + 3(3 + 3t) + 5(5t – 5) = – 2 i.e., when t = 
Substituting this value of t in (2), we get 
= 
+ 
= 
∴ The reqd. point of intersection is 
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