The vertices B and C of a `DeltaABC` lie on the line, `(x+2)/(3)=(y-1)/(0)=(z)/(4)` such that BC=5 units. Then, the area (in sq units) of this triangle, given that the point A(1, -1, 2) is

Options

A. `sqrt(34)`
B. `2sqrt(34)`
C. `5sqrt(17)`
D. 6

Answer

A

Explanation

Given line is `(x+2)/(3)=(y-1)/(0)=(z)/(4)`
Vector along line is, a `=3hati+4hatk`
and vector joining the points (1, -1, 2) to (-2, 1, 0) is
`b=(1+2)hati+(-1-1)hatj+(2-0)hatk`
`=3hatj-2hatj+2hatk`
and `|BC|=5` units
Now, area of required `DeltaABC`
`=(1)/(2)|BC||b||sintheta|" "...(ii)`
[where `theta` is angle between vectors a and b]
`because|b|sintheta=(|axxb|)/(|a|),`
`because|axxb|=|{:(hati," "hatj," "hatk),(3," "0," "4),(3,-2," "2):}|=8hati+6hatj-6hatk`
`implies|axxb|=sqrt(64+36+36)`
`=sqrt(136)=2sqrt(
34)`
and `" "|a|=sqrt(9+16)=5`
`:.|b|sintheta=(2sqrt(34))/(5)`
On substituting these values in Eq. (i), we get
Required area `=(1)/(2)xx5xx(2sqrt(34))/(5)=sqrt(34)`sq units
Alternate Method
Given line is `(x+2)/(3)=(y-1)/(0)=(z)/(4)=lambda (let)" "...(i)`

Since, point D lies on the line BC.
`:."Coordinates of "D=(3lambda-2,1,4lambda)`
Now,`" "DR" of "BCimpliesa_(1)=3, b_(1)=0, c_(1)=4`
and `DR" of "ADimpliesa_(2)=3lambda-3, b_(2)=2, c_(2)=4lambda-2`
Since, `AD_|_BC,a_(1)a_(2)+b_(1)b_(2)+c_(1)c_(2)=0`
`3xx(3lambda-3)+0(2)+4(4lambda-2)=0`
`implies9lambda-9+0+16lambda-8=0`
`implies25lambda-17=0`
`implies" "lambda=(17)/(25)`
`:."Coordinates of "D=((1)/(25),1,(68)/(25)).`
Now, `AD=sqrt((1-(1)/(25))^(2)+(-1-1)^(2)+(2-(68)/(25))^(2))`
`=sqrt(((24)/(25))^(2)+(-2)^(2)+((-18)/(25))^(2))`
`=sqrt((576)/(625)+4+(324)/(625))=(2)/(5)sqrt(34)`
`:."Area of "DeltaABC=(1)/(2)BCxxAD`
`=(1)/(2)xx5xx(2)/(5)sqrt(34)`
`=sqrt(34)`sq units