Question

Find the value of $\lambda $ for which the points $(6, - 1,2)$, $(8, - 7,\lambda )$ and $(5,2,4)$ are collinear.

Answer 1

Hint: This problem comes under co–ordinate geometry. We need to use section formulas internally to solve this problem. The points are collinear which means all points lie on the same line. Since the points are collinear initially we will solve \[l:1\] ratios and then we get the ratio value of \[l\] with that we can find $\lambda $ and then basic mathematical calculation and complete step by step explanation.

Formula used: Section formula internally in the ratio \[l:m\]
\[(x,y) = \left( {\dfrac{{l{x_2} + m{x_1}}}{{l + m}},\dfrac{{ly{}_2 + m{y_2}}}{{l + m}}} \right)\]

Complete step-by-step answer:
Consider the given collinear points as $A(6, - 1,2)$, $B(8, - 7,\lambda )$, $C(5,2,4)$
Let us assume
$A(6, - 1,2) = \left( {{x_1},{y_1},{z_1}} \right)$
$B(8, - 7,\lambda ) = \left( {{x_2},{y_2},{z_2}} \right)$
$C(5,2,4) = \left( {{x_3},{y_3},{z_3}} \right)$
Use the formula mentioned in formula used and take only $x$ – axis coordinates only
\[ \Rightarrow x = \dfrac{{l{x_2} + m{x_1}}}{{l + m}}\]
Set \[B\] cuts \[AC\] in the ratio \[l:1\]and then substitute the value of $x$ on their respective places
\[ \Rightarrow {x_2} = \dfrac{{l{x_3} + m{x_1}}}{{l + m}}\]
Substituting the values ${x_3} = 5$ and ${x_1} = 6$,
\[ \Rightarrow 8 = \dfrac{{l(5) + 1(6)}}{{l + 1}}\]
Simplifying we get,
\[ \Rightarrow 8 = \dfrac{{5l + 6}}{{l + 1}}\]
Now re-arrange the values, we get
\[ \Rightarrow 8(l + 1) = 5l + 6\]
Multiplying the terms,
\[ \Rightarrow 8l + 8 = 5l + 6\]
Now separate \[l\],
\[8l - 5l = 6 - 8\]
Solving for \[l\] we get,
\[ \Rightarrow 3l = - 2\]
\[l = \dfrac{{ - 2}}{3}\]
Hence, \[B\] cuts \[AC\] in the ratio \[\dfrac{{ - 2}}{3}:1\]
Similarly, Use the formula mentioned in formula used and take only \[z\] – axis co-ordinates only because we need to find \[\lambda \]
\[ \Rightarrow {z_2} = \dfrac{{l{z_3} + m{z_1}}}{{l + m}}\]
Substituting the values,
\[ \Rightarrow \lambda = \dfrac{{\left( {\dfrac{{ - 2}}{3}} \right)(4) + (2)}}{{\left( {\dfrac{{ - 2}}{3}} \right) + 1}}\]
Simplifying we get,
\[ \Rightarrow \lambda = \dfrac{{\dfrac{{ - 8}}{3} + 2}}{{\dfrac{{ - 2}}{3} + 1}}\]
Now, we are going to do fraction addition, we get
\[ \Rightarrow \lambda = \dfrac{{\dfrac{{ - 8}}{3} + \dfrac{6}{3}}}{{\dfrac{{ - 2}}{3} + \dfrac{3}{3}}}\]
The denominators are same, so simplifying we get,
\[ \Rightarrow \lambda = \dfrac{{\dfrac{{ - 8 + 6}}{3}}}{{\dfrac{{ - 2 + 3}}{3}}}\]
Adding the terms we get,
\[ \Rightarrow \lambda = \dfrac{{\dfrac{{ - 2}}{3}}}{{\dfrac{1}{3}}}\]
Taking reciprocal to solve,
\[ \Rightarrow \lambda = \dfrac{{ - 2}}{3} \times \dfrac{3}{1}\]
By cancelling \[3\], we get
\[\lambda = - 2\]

$\therefore $ Thus the value of \[\lambda \] is \[ - 2\].

Note: This kind of problem needs attention on the section formula. We need to know about the collinear points so that we can find the ratio. For that we are able to solve the equation in similar methods. First we find the ratio of \[l\] and then with that again using the section formula we find \[\lambda \]. Like this kind of problem to find a missing point we have to use this kind of method which is easy to solve and makes the solution simple.