If (1, 5, 35), (7, 5, 5), (1, $\lambda$, 7) and (2$\lambda$, 1, 2) are coplanar, then the sum of all possible values of $\lambda$ is :
Solution
A(1, 5, 35), B(7, 5, 5), C(1, $\lambda$, 7), D(2$\lambda$, 1, 2)<br><br>$\overrightarrow {AB}$ = 6$\widehat i$ $-$ 30$\widehat k$,
<br><br>$\overrightarrow {BC}$ = $-$6$\widehat i$ ($\lambda$ $-$ 5)$\widehat j$ + 2$\widehat k$,
<br><br>$\overrightarrow {CD}$ = (2$\lambda$ $-$ 1)$\widehat i$ + (1 $-$ $\lambda$)$\widehat j$ $-$ 5$\widehat k$<br><br>Points are coplanar<br><br>$$ \Rightarrow 0 = \left| {\matrix{
6 & 0 & { - 30} \cr
{ - 6} & {\lambda - 5} & 2 \cr
{2\lambda - 1} & {1 - \lambda } & { - 5} \cr
} } \right|$$<br><br>= 6($-$5$\lambda$ + 25 $-$ 2 + 2$\lambda$) $-$ 30($-$6 + 6$\lambda$ $-$ (2$\lambda$<sup>2</sup> $-$ $\lambda$ $-$ 10$\lambda$ + 5))<br><br>= 6($-$3$\lambda$ + 23) $-$ 30($-$2$\lambda$<sup>2</sup> + 11$\lambda$ $-$ 5 $-$ 6 + 6$\lambda$)<br><br>= 6($-$3$\lambda$ + 23) $-$ 30($-$2$\lambda$<sup>2</sup> + 17$\lambda$ $-$11)<br><br>= 6($-$3$\lambda$ + 23 + 10$\lambda$<sup>2</sup> $-$ 85$\lambda$ + 55)<br><br>= 6(10$\lambda$<sup>2</sup> $-$ 88$\lambda$ + 78) = 12(5$\lambda$<sup>2</sup> $-$ 44$\lambda$ + 39)<br><br>$\Rightarrow$ 0 = 12(5$\lambda$<sup>2</sup> $-$ 44$\lambda$ + 39)
<br><br>$\Rightarrow$ 5$\lambda$<sup>2</sup> $-$ 44$\lambda$ + 39 = 0
<br><br>this quadratic equation has two values $\lambda$<sub>1</sub> and $\lambda$<sub>2</sub>
<br><br>$\therefore$ $\lambda$<sub>1</sub> + $\lambda$<sub>2</sub> = ${{44} \over 5}$
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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