Q. 285.0( 2 Votes )

Find the coordina

Answer :

Given: a point P (5, 4, 2), line .


To find: the length and coordinates of the foot of the perpendicular from the point P to the given line. Also find the image of point P in the given line.


Given line is



So the Cartesian form of the line will be,



Let Q(α, β, γ) be the foot of perpendicular drawn from P(5, 4, 2) to the line (i) and P’ (x1 , y1 , z1) be the image of P on the line (i)



Now as point Q lies on the given line, so let



So from above condition we get





Hence the coordinates of Q are (2k-1, 3k+3, 1-k)


Now


Parallel vector of the given line is


Obviously,



Substituting corresponding values, we get


2(α-5)+3(β-4)+(-1)(γ-2)=0


2α-10+3β-12-γ+2=0


2α+3β-γ-20=0


Now substituting the values of α, β, γ, we get


2(2k-1)+3(3k+3)-(1-k)-20=0


4k-2+9k+9-1+k-20=0


14k-14=0


14k=14


k=1


Substituting the value of k in coordinates of Q, we get


Q(2k-1, 3k+3, 1-k)


Q(2(1)-1, 3(1)+3, 1-(1))


Q(2-1, 3+3, 1-1)


Q(1, 6, 0)


Hence coordinates of the foot of perpendicular and the length of the perpendicular drawn from the point P (5, 4, 2) to the line is Q (1,6,0)


Now we will find the length of this perpendicular


So





Or the length of the perpendicular is √24 or 2√6 units.


Now Let P’(x1, y1, z1) be the image of point P (5, 4, 2) in the line


Now from figure it is clear that M(1, 6, 0) is the midpoint of line PP’


Hence now applying the formula





Hence the image of the given point P(5, 4, 2) is P’(-3, 8, -2).


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