A particle just clears a wall of height b at distance a and strikes the ground at a distance c from the point of projection. The angle of projection is (1) `tan^(-1)b/(a c)` (2) `"45"^o` (3) `tan^(-1)(b c)/(a(c-a)` (4) `tan^(-1)(b c)/a`

Options

A. `(1)/(a^(2))+(1)/(b^(2))+(1)/(c^(2))+(1)/(a'^(2))+(1)/(b'^(2))+(1)/(c'^(2))=0`
B. `(1)/(a^(2))-(1)/(b^(2))-(1)/(c^(2))-(1)/(a'^(2))-(1)/(b'^(2))-(1)/(c'^(2))=0`
C. `(1)/(a^(2))+(1)/(b^(2))+(1)/(c^(2))-(1)/(a'^(2))-(1)/(b'^(2))-(1)/(c'^(2))=0`
D. `(1)/(a^(2))+(1)/(b^(2))+(1)/(c^(2))+(1)/(a'^(2))+(1)/(b'^(2))+(1)/(c'^(2))=0`

Answer

c

Explanation

The planes are `(x)/(a)+(y)/(b)+(z)/(c)=1and(x)/(a')+(y)/(b')+(z)/(c')=1`
Since the perpendicular distaance of the origion on the planes is same, therefore,
`|(-1)/(sqrt((1)/a^(2)+(1)/(b^(2))+(1)/(c^(2))))|=|(-1)/(sqrt((1)/(a'^(2))+(1)/(b'^(2))+(1)/(c'^(2))))|`
or `(1)/(a^(2))+(1)/(b^(2))+(1)/(c^(2))-(1)/(a'^(2))-(1)/(b'^(2))-(1)/(c'^(2))=0`