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 )/3
Also, 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: \frac{(4 - x)}{2} = \frac{y}{6} = \frac{(1 - z)}{3}

This line can also be written as: \frac{(X - 4)}{- 2} = \frac{y}{6} = \frac{(z - 1)}{- 3}

Next, let us consider \frac{(X - 4)}{- 2} = \frac{y}{6} = \frac{(z - 1)}{- 3} = 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: (X2 – X1, Y2 – Y1, Z2 – Z1)

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

a1 a2 + b1 b2 + c1 c2= 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 = \frac{49}{49}

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 \sqrt{(X_2 - X_1)^2 + (Y_2 - Y_1)^2 + (Z_2 - Z_1)^2}

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

= \sqrt{(2 - 2)^2 + (6 - 3)^2 + (- 2 - (- 8))^2}

= \sqrt{0 + 9 + 36}

= \sqrt 45 = 3 \sqrt 5 units

Therefore, the required coordinates of the foot of perpendicular are (2, 6, – 2) and the required distance is 3\sqrt5 units.

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