Equation of plane :
To Find :
Equation of line
Point of intersection
Equation of line passing through point A with position vector and parallel to vector is
From the given equation of the plane
The normal vector of the plane is
As the given line is perpendicular to the plane therefore will be parallel to the line.
Now, the equation of the line passing through and parallel to is
This is the required equation line.
Substituting in eq(1)
⇒6x – 3y + 5z = -2 ………eq(3)
Also substituting in eq(2)
Comparing coefficients of
Let Q(a, b, c) be the point of intersection of given line and plane
As point Q lies on the given line.
Therefore from eq(4)
⇒a = 6k+2, b = -3k-3, c = 5k-5
Also point Q lies on the plane.
Therefore from eq(3)
6a – 3b + 5c = -2
⇒6(6k+2) – 3(-3k-3) + 5(5k-5) = -2
⇒36k + 12 + 9k + 9 + 25k – 25 = -2
⇒70k = 2
Therefore co-ordinates of the point of intersection of line and plane are
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