**Q) Find the co-ordinates of the foot of the perpendicular drawn from the point (2, 3, – 8) to the line (4 – x)/2 = y/6 = (1 – z )/3Also, find the perpendicular distance of the given point from the line.**

**On the basis of the above information, answer the following questions :****(i) Write the expression for the area of the visiting card in terms of x.****(ii) Obtain the dimensions of the card of minimum area.**

**Ans: **Let P be the point (2, 3,- 8) and Q be the be the foot of perpendicular from the point P on line

Q is the general point on the line:

This line can also be written as:

Next, let us consider = λ

∴ Coordinates of Q = (4 – 2 λ, 6 λ, 1 – 3 λ)

Next, we know that the We know the direction ratios of any line segment are given by: (X_{2} – X_{1}, Y_{2} – Y_{1}, Z_{2} – Z_{1})

Therefore, direction ratios of line PQ = (4 – 2 λ – 2), (6 λ – 3), (1 – 3 λ – (- 8))

= (2 – 2λ, 6 λ – 3, 9 – 3 λ)

Since PQ is perpendicular to the given line, hence the sum of the product of this direction ratios = 0

a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}= 0

(- 2) (2 – 2 λ) + (6) (6 λ – 3) + (- 3) (9 – 3 λ) = 0

– 4 + 4 λ + 36 λ – 18 – 27 + 9 λ = 0

49 λ – 49 = 0

λ =

λ = 1

Hence, the coordinates of Q are: (4 -2 λ, 6 λ, 1 – 3 λ)

= (4 – 2 (1), 6 (1), 1 – 3 (1))

= (2, 6,- 2)

Next, we know the distance between two points is given by

Perpendicular distance PQ between P (2, 3, – 8) and Q (2, 6,- 2)

=

=

= units

**Therefore, the required coordinates of the foot of perpendicular are (2, 6, – 2) and the required distance is units.**

**Please press Heart if you liked the solution.**