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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 :

JEE · Math · previous-year question

  1. A.$$ - {{44} \over 5}$$
  2. B.$$ - {{39} \over 5}$$
  3. C.$${{44} \over 5}$$correct
  4. D.$${{39} \over 5}$$

Answer

C. $${{44} \over 5}$$

Explanation

A(1, 5, 35), B(7, 5, 5), C(1, $$\lambda$$, 7), D(2$$\lambda$$, 1, 2) $$\overrightarrow {AB} $$ = 6$$\widehat i$$ $$-$$ 30$$\widehat k$$, $$\overrightarrow {BC} $$ = $$-$$6$$\widehat i$$ ($$\lambda$$ $$-$$ 5)$$\widehat j$$ + 2$$\widehat k$$, $$\overrightarrow {CD} $$ = (2$$\lambda$$ $$-$$ 1)$$\widehat i$$ + (1 $$-$$ $$\lambda$$)$$\widehat j$$ $$-$$ 5$$\widehat k$$ Points are coplanar $$ \Rightarrow 0 = \left| {\begin{matrix} 6 & 0 & { - 30} \\ { - 6} & {\lambda - 5} & 2 \\ {2\lambda - 1} & {1 - \lambda } & { - 5} \\ \end{matrix} } \right|$$ = 6($$-$$5$$\lambda$$ + 25 $$-$$ 2 + 2$$\lambda$$) $$-$$ 30($$-$$6 + 6$$\lambda$$ $$-$$ (2$$\lambda$$2 $$-$$ $$\lambda$$ $$-$$ 10$$\lambda$$ + 5)) = 6($$-$$3$$\lambda$$ + 23) $$-$$ 30($$-$$2$$\lambda$$2 + 11$$\lambda$$ $$-$$ 5 $$-$$ 6 + 6$$\lambda$$) = 6($$-$$3$$\lambda$$ + 23) $$-$$ 30($$-$$2$$\lambda$$2 + 17$$\lambda$$ $$-$$11) = 6($$-$$3$$\lambda$$ + 23 + 10$$\lambda$$2 $$-$$ 85$$\lambda$$ + 55) = 6(10$$\lambda$$2 $$-$$ 88$$\lambda$$ + 78) = 12(5$$\lambda$$2 $$-$$ 44$$\lambda$$ + 39) $$ \Rightarrow $$ 0 = 12(5$$\lambda$$2 $$-$$ 44$$\lambda$$ + 39) $$ \Rightarrow $$ 5$$\lambda$$2 $$-$$ 44$$\lambda$$ + 39 = 0 this quadratic equation has two values $$\lambda$$1 and $$\lambda$$2 $$ \therefore $$ $$\lambda$$1 + $$\lambda$$2 = $${{44} \over 5}$$

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