If (x, y, z) be an arbitrary point lying on a plane P which passes through the points (42, 0, 0), (0, 42, 0) and (0, 0, 42), then the value of the expression
$$3 + {{x - 11} \over {{{(y - 19)}^2}{{(z - 12)}^2}}} + {{y - 19} \over {{{(x - 11)}^2}{{(z - 12)}^2}}} + {{z - 12} \over {{{(x - 11)}^2}{{(y - 19)}^2}}} - {{x + y + z} \over {14(x - 11)(y - 19)(z - 12)}}$$ is equal to :
Solution
From intercept from, equation of plane is x + y + z = 42<br><br>$\Rightarrow$ (x $-$ 11) + (y $-$ 19) + (z $-$ 12) = 0<br><br>let a = x $-$ 11, b = y $-$ 19, c = z $-$ 12<br><br>a + b + c = 0<br><br>Now, given expression is<br><br>$$3 + {a \over {{b^2}{c^2}}} + {b \over {{a^2}{c^2}}} + {c \over {{a^2}{b^2}}} - {{42} \over {14abc}}$$<br><br>$3 + {{{a^3} + {b^3} + {c^3} - 3abc} \over {{a^2}{b^2}{c^2}}}$<br><br>If a + b + c = 0<br><br>$\Rightarrow$ a<sup>3</sup> + b<sup>3</sup> + c<sup>3</sup> = 3 abc<br><br>$\therefore$ $3 + {0 \over {{a^2}{b^2}{c^2}}}$<br><br>$= 3$
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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