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If the points `(2-x, 2, 2), (2, 2-y, 2) and (2, 2, 2-z)` are coplanar then prove `2/x+2/y+2/z=1`A. `1/x+1/y+1/z=1`B. `x+y+z=1`C. `1/(1-x)+1/(1-y)+1/(1-z)=1`D. `1/(1-x)+1/(1-y)+1/(1-z)=2` |
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Answer» Correct Answer - A `vec(AB)=vec(OB)-vec(OA)={2hati+(2-y)hatj+2hatk}-` `{(2-x)hati+2hatj+2hatk}=xhati-yhatj` `vec(AC)=vec(OC)-vec(OA)` `={(2hati+2hatj+(2-z)hatk}-{(2-x)hati+2hatj+2hatk}` `=xhati-zhatk` `vec(AD)=vec(OD)-vec(OA)` `=(hati+hatj+hatk)-{(2-x)hati+2hatj+2hatk}=(x-1)hati-hatj-hatk` As these vectors are coplanar, `|(x, -y, 0),(x, 0, -z),(x-1, -1, -1)|=0` On simplication we get `1/x+1/y+1/z=1` |
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