Question

Given that B and C lie on the line `(x + 3)/5 = (y - 1)/2 = (z + 4)/3` and BC = 5 units, find the area of ΔABC.

Answer 1

B and C lie on the line `(x + 3)/5 = (y - 1)/2 = (z + 4)/3`

So direction ratios of line BC are (5, 2, 3).

BC = 5 units

From the diagram, AD ⊥ BC (D is the foot of the perpendicular from A on line BC).

A(0, 2, 3) (as used in the solution).

Let `(x + 3)/5 = (y - 1)/2 = (z + 4)/3 = λ`

x = 5λ − 3, y = 2λ + 1, z = 3λ − 4

So any point D on BC is D(5λ – 3, 2λ + 1, 3λ – 4).

Use the perpendicular condition AD ⊥ BC:

`vec(AD) = vecD - vecA`

= `(5λ - 3 - 0)hati + (2λ + 1 - 2)hatj + (3λ - 4 - 3)hatk`

= `(5λ - 3)hati + (2λ - 1)hatj + (3λ - 7)hatk`

Direction vector of line BC is:

`Vec(BC) = 5hati + 2hatj + 3hatk`

Since AD ⊥ BC,

`vec(AD)*vec(BC) = 0`

(5λ − 3) (5) + (2λ − 1) (2) + (3λ − 7) (3) = 0

25λ − 15 + 4λ − 2 + 9λ − 21 = 0

38λ − 38 = 0

λ = 1

∴ D[5(1) − 3, 2(1) + 1, 3(1) − 4]

= (2, 3, −1)

Coordinates of D are (2, 3, –1).

`AD = sqrt((2 - 0)^2 + (3 - 2)^2 + (-1 - 3)^2`

= `sqrt(4 + 1 + 16)`

= `sqrt(21)`

Area of ∆ABC = `1/2 xx BC xx AD`

= `1/2 xx 5 xx sqrt(21)`

= `5/2 sqrt(21)` sq. units