Question

The plane $XOZ$ divides the join of $\left( {1, - 1,5} \right){\text{ and }}\left( {2,3,4} \right)$ in the ratio $\lambda:1$, then $\lambda $ is ?
$
  {\text{A}}{\text{. - 3}} \\
  {\text{B}}{\text{. }}\dfrac{{ - 1}}{3} \\
  {\text{C}}{\text{. 3}} \\
  {\text{D}}{\text{. }}\dfrac{1}{3} \\
 $

Answer 1

Hint: We will draw the figure and by using the figure we acknowledge the point where the plane cuts the line. Then, we will use section formula and find the value of $\lambda $ and match that particular value with these options mentioned above.

Complete step-by-step answer:
Given points are $\left( {1, - 1,5} \right)$ and $\left( {2,3,4} \right)$ and a plane$XOZ$ where, $X$ refers to $x - plane$, $O$ refers to $origin$ and $Z$ refers to $z - plane$ which cuts the line segment. Thus, the point is in $XZ$$plane$ and cuts the line in a ratio of $\lambda :1$.

Using section formula i.e.
If the coordinates of the line $\left( {{x_1},{y_1},{z_1}} \right)$ and $\left( {{x_2},{y_2},{z_2}} \right)$ are divided in the ratio \[m:n\] then desired points will be $\left( {\dfrac{{n{x_1} + m{x_2}}}{{n + m}},\dfrac{{n{y_1} + m{y_2}}}{{n + m}},\dfrac{{n{z_1} + m{z_2}}}{{n + m}}} \right)$.
We can say, the desired points in our case will be $\left( {\dfrac{{2\lambda + 1}}{{\lambda + 1}},\dfrac{{3\lambda - 1}}{{\lambda + 1}},\dfrac{{4\lambda + 5}}{{\lambda + 1}}} \right)$
Since, this point then lies in $XOZ$$plane$ then it’s $y - coordinate$ should be zero.
$\therefore {\text{ }}\dfrac{{3\lambda - 1}}{{\lambda + 1}} = 0$
We get
$
  3\lambda - 1 = 0 \\
   \Rightarrow 3\lambda = 1 \\
  \lambda = \dfrac{1}{3} \\
 $
Hence, the correct option is ${\text{D}}{\text{. }}\dfrac{1}{3}$

Note: In the 3-D geometry we’ll use a section formula which tells us the coordinates of the points which divides a given line segment into two parts such that their lengths are in the ratio $m:n$. Do note that section formula is very helpful in coordinate geometry and in this question. Thus, this is the only way.