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Lines L(1):(x-6)/3=(y-4)/2=z-2 and L(2):...

Lines `L_(1):(x-6)/3=(y-4)/2=z-2` and `L_(2):(x-8)/4=y-2=(z-4)/2` meets the plane `pi:vecr(2hati+hatj-hatk)=7` at points A and B
(2)Area of the triangle formed by the lines `L_(1),L_(2)` & `AB` is

A

`7/2 sqrt(19/2)`

B

`sqrt(19/2)`

C

`(sqrt(11))/4`

D

`6sqrt(5)`

Text Solution

Verified by Experts

The correct Answer is:
B
`A-=(3,2,1), B-=(4,1,2)` point of intersection `-=(0,0,0)`
Area of triangle `=1/2|(3hati+2hatj+hatk)xx(4hati+hatj+2hatk)|=sqrt(19/2)`
Also image of `(0,0,0)` w.r.t plane `pi` is `(14/3, 7/3, (-7)/3)`
Image of `(0,0,0)` w.r.t `AB` is `(14/3, 16/3, 2/3)`
Equation of line joining these two points `=(3x-14)/0=(3y-7)/3=(3z+7)/3`
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