Question

The line joining the points (2, -3, 1) and (3, -4, -5) cuts the coordinate plane at the point.
This question has multiple correct options.
$
  {\text{A}}{\text{. }}\left( {0, - 1,13} \right) \\
  {\text{B}}{\text{. }}\left( {0,0,1} \right) \\
  {\text{C}}{\text{. }}\left( { - 1,0,19} \right) \\
  {\text{D}}{\text{. }}\left( {8, - 9,0} \right) \\
$

Answer 1

Hint: In order to find the point where the joining the points (2, -3, 1) and (3, -4, -5) cuts the coordinate plane, we have to determine the line equation which describes each of the coordinate x, y and z respectively. Then we determine where the point x cuts the yz plane, point y cuts the xz plane and the point z cuts the xy plane respectively.

Complete step by step solution:
Given data,
Points are (2, -3, 1) and (3, -4, -5)

We know, the equation of a line joining two points of the form (a, b, c) and (p, q, r) is given by the formula
$\dfrac{{{\text{x - a}}}}{{{\text{a - p}}}} = \dfrac{{{\text{y - b}}}}{{{\text{b - q}}}} = \dfrac{{{\text{z - c}}}}{{{\text{c - r}}}} = {\text{k}}$, where k is a constant.
Now the equation of a line passing through the points (2, -3, 1) and (3, -4, -5) is given by,
Let us take the constant as some variable ‘r’
$ \Rightarrow \dfrac{{{\text{x - 2}}}}{{{\text{2 - 3}}}} = \dfrac{{{\text{y + 3}}}}{{{\text{ - 3 + 4}}}} = \dfrac{{{\text{z - 1}}}}{{{\text{1 + 5}}}} = {\text{r}}$
$ \Rightarrow \dfrac{{{\text{x - 2}}}}{{{\text{ - 1}}}} = \dfrac{{{\text{y + 3}}}}{1} = \dfrac{{{\text{z - 1}}}}{6} = {\text{r}}$
⟹x = -r + 2, y = r – 3, z = 6r +1
Therefore the line containing the point (-r + 2, r – 3, 6r + 1) cuts the yz plane at x = 0, xy plane at z = 0 and zx plane at y = 0 respectively.
And the point of intersection for each plane can be found by substituting the value of r obtained by substituting the respective coordinate equal to 0.
Therefore the line containing the point (-r + 2, r – 3, 6r + 1) cuts the yz plane at x = 0
⟹-r + 2 = 0
⟹r = 2
The point of intersection is, (-r + 2, r – 3, 6r + 1) = (-2 + 2, 2 – 3, 6(2) + 1) = (0, -1, 13).

Therefore the line containing the point (-r + 2, r – 3, 6r + 1) cuts the xy plane at z = 0
⟹6r + 1 = 0
⟹r = $ - \dfrac{1}{6}$
The point of intersection is, (-r + 2, r – 3, 6r + 1) = ($\dfrac{1}{6}$ + 2, $ - \dfrac{1}{6}$ – 3, 6$\left( { - \dfrac{1}{6}} \right)$ + 1) = ($\dfrac{{13}}{6}$,$ - \dfrac{{19}}{6}$, 0).

Therefore the line containing the point (-r + 2, r – 3, 6r + 1) cuts the zx plane at y = 0
⟹r - 3 = 0
⟹r = 3
The point of intersection is, (-r + 2, r – 3, 6r + 1) = (-3 + 2, 3 – 3, 6(3) + 1) = (-1, 0, 19).

Hence the points of intersection are (0, -1, 13), ($\dfrac{{13}}{6}$,$ - \dfrac{{19}}{6}$, 0) and (-1, 0, 19).
Hence options A and C are the correct answers.

Note: In order to solve this type of questions the key is to know the formula of a general equation passing through a point, in 3-dimensional space. It is important to know that in a plane which is two dimensional only two coordinates of the given point exist, one coordinate has to be zero. We used this concept to determine the x, y and z coordinates while solving the line equation. Then by just substituting the values of x, y and z we get the points of intersection of their respective planes.