devarshi-dt-logo

Question:

The vertices B and C of a triangle ABC 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:

√34

6

5√17

2√34

Solution:

The correct option is B
√34
AD (3^i+4^k)=0
3(3λ-3)+0+4(4λ-2)=0
(9λ-9)+(16λ-8)=0
25λ=17
λ=17/25
AD=(51/25-3)^i+2^j+(68/25-2)^k=24/25^i+2^j+18/25^k
AD = √576/625+4+324/625=√3400/625=25√34
Area of triangle=Δ=1/2 × 5 × 2√34/5=√34