Question 476 of 521MediumMCQ
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 :
Solutions · 1
Official solution
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
\[ \Rightarrow 0 = \left| {6−62λ−10λ−51−λ−302−5 } \right|\]
= 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 = \[{{44} \over 5}\] Answer the question to unlock