If the plane `x/2+y/3+z/6=1` cuts the axes of coordinates at points, `A ,B, ` and `C`, then find the area of the triangle `A B C`. a. `18` sq unit b. `36` sq unit c. `3sqrt(14)` sq unit d. `2sqrt(14)` sq unit

Options

A. `sqrt(29)` sq. units
B. `sqrt(41)` sq. units
C. `sqrt(61)` sq. units
D. none of these

Answer

C

Explanation

The given plane cuts the coordinate axes in `A(2,0,0),B(0,3,0)` and `C(0,0,4)`.
`:.` Area of `DeltaABC=1/2ABxxACxxsin/_BAC`
Now,
`AB=sqrt(4+9+0)=sqrt(13),AC=sqrt(4+0+16)=sqrt(20)`
`cos /_BCA(vec(AB).vec(AC))/(|vec(AB)||vec(AC)|)=((-2hati+3hatj).(-2hati+4hatk))/(sqrt(4+9)sqrt(4+16))`
`impliescos /_BAC=(4+0+0)/(sqrt(13)sqrt(20))=4/(sqrt(13)sqrt(20))=2/(sqrt(65))`
`impliessin /_BAC=sqrt(1-4/65)=sqrt(61/65)`
Hence
Area of `DeltaABC=1/2xxsqrt(13)xxsqrt(20)xxsqrt(61/65)=sqrt(61)` sq. units.