Q. 185.0( 1 Vote )

Find the length and the foot of perpendicular from the point to the plane 2x – 2y + 4z + 5 = 0.

Answer :

It is given that,

The point is .


The plane is 2x – 2y + 4z + 5 = 0


We need to find the foot of the perpendicular from the Point P to the equation of the given plane.


Also, we need to find the distance from the point P to the plane.


Let Q be the foot of the perpendicular from the point .


Let the point Q be Q(x1, y1, z1).


We know the direction ratio of any line segment PQ, where P(x1, y1, z1) and Q(x2, y2, z2), is given by (x2 – x1, y2 – y1, z2 – z1).


So, the direction ratio of PQ is given by


Direction ratio of PQ = (x1 – 1, y1 – 3/2, z1 – 2)


Now, let be the normal to the plane 2x – 2y + 4z + 5 = 0.


is obviously parallel to the since, a normal is an object such as a line or vector that is perpendicular to a given object.


Direction ratio simply states the number of units to move along each axis.


For any plane, ax + by + cz = d


a, b and c are normal vectors to the plane.


And therefore, the direction ratios are (a, b, c).


So, direction ratio of = (2, -2, 4) for plane 2x – 2y + 4z + 5 = 0.


Cartesian equation of line PQ, where P(1, 3/2, 2) and Q(x1, y1, z1) is



Let us find any point on this line.



From


x1 – 1 = 2λ


x1 = 2λ + 1


From




From


z1 – 2 = 4λ


z1 = 4λ + 2


Any point on the line is (2λ + 1, 3/2 – 2λ, 4λ + 2).


This point is Q.


Q(x1, y1, z1) = Q(2λ + 1, 3/2 – 2λ, 4λ + 2) …(i)


And it was assumed to be lying on the given plane. So, substitute x1, y1 and z1 in the plane equation, we get


2x1 – 2y1 + 4z1 + 5 = 0



Simplifying it to find the value of λ.


4λ + 2 – 3 + 4λ + 16λ + 8 + 5 = 0


4λ + 4λ + 16λ + 2 – 3 + 8 + 5 = 0


24λ + 12 = 0


24λ = -12




Since, Q is the foot of the perpendicular from the point P.


The, substitute the value of λ in equation (i),





Also, we need to find .


Where,


P = (1, 3/2, 2)


Q = (0, 5/2, 0)


is found as,






Thus, foot of the perpendicular from the given point to the plane is (0, 5/2,0) and the distance is √6 units.


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