Question

Find the coordinate of the point where the line through the points $(3, - 4, - 5)$ and $(2, - 3,1)$ cross the plane determined by the $(1,2,3)$ , $(4,2, - 3)$ and $(0,4,3)$ .

Answer 1

Hint: The coordinate of the point is the place from where the line passes to meet the stated condition. A plane is a two-dimensional analogue of a point, a line and three-dimensional space and a normal plane a vector perpendicular to the tangent plane of the square at $P$ .

Completed step-by-step solution:
Given: The coordinates of line $(3, - 4, - 5)$ and $(2, - 3,1)$, plane $(1,2,3)$ , $(4,2, - 3)$ and $(0,4,3)$ .
We know that the Cartesian equation of a line passing through two points $({x_1},{y_1},{z_1})$ and $({x_2},{y_2},{z_2})$ is given by
$\dfrac{{x - {x_1}}}{{{x_2} - {x_1}}} = \dfrac{{y - {y_1}}}{{{y_2} - {y_1}}} = \dfrac{{z - {z_1}}}{{{z_2} - {z_1}}}$
So, the question of a line passing through $(3, - 4, - 5)$ and $(2, - 3,1)$ is
$
  \dfrac{{x - 3}}{{2 - 3}} = \dfrac{{y - ( - 4)}}{{(-3) - ( - 4)}} = \dfrac{{z - ( - 5)}}{{1 - ( - 5)}} \\
   \Rightarrow \dfrac{{x - 3}}{{ - 1}} = \dfrac{{y - ( - 4)}}{1} = \dfrac{{z - ( - 5)}}{6} \\
   \Rightarrow \dfrac{{x - 3}}{{ - 1}} = \dfrac{{y + 4}}{1} = \dfrac{{z + 5}}{6} \\
 $
Now, coordinates of any points on this line are given by
$\dfrac{{x - 3}}{{ - 1}} = \dfrac{{y + 4}}{1} = \dfrac{{z + 5}}{6} = k$
$ \Rightarrow x = 3 - k,y = k - 4,z = 6k - 5,$ where $k$ is a constant.
Let $R(3 - k,k - 4,6k - 5)$ be the required point of intersection.
Now, let the equation of a plane passing through $(1,2,3)$ be
$a(x - 1) + b(y - 2) + c(z - 3) = 0 \to (1)$
Here $a,b,c$ are the direction ratios of a normal to the plane.
Since the plane (1) passes through $(4,2, - 3)$ , so
$
  a(4 - 1) + b(2 - 2) + c( - 3 - 3) = 0 \\
   \Rightarrow 3a - 6c = 0 \to (2) \\
 $
Also, the plane passes through $(0,4,3)$ so
$
  a(0 - 1) + b(4 - 2) + c(3 - 3) = 0 \\
   \Rightarrow - a + 2b = 0 \to (3) \\
 $
Solving equation (2) and (3) using the method of cross- multiplication we have,
$
  \dfrac{a}{{0 + 12}} = \dfrac{b}{{6 - 0}} = \dfrac{c}{{6 + 0}} \\
   \Rightarrow \dfrac{a}{{12}} = \dfrac{b}{6} = \dfrac{c}{6} \\
   \Rightarrow \dfrac{a}{2} = b = c = \lambda (say) \\
   \Rightarrow a = 2\lambda ,b = \lambda ,c = \lambda . \\
 $
Now putting the value of equation (1) we get
$
  2\lambda (x - 1) + \lambda (y - 2) + \lambda (z + 3) = 0 \\
   \Rightarrow 2x + y + z - 7 = 0 \to (4) \\
 $
Putting $x = 3 - k,y = k - 4,z = 6k - 5$ in equation (4) we get
$
   \Rightarrow 2(3 - k) + (k - 4) + (6x - 5) - 7 = 0 \\
   \Rightarrow 5k - 10 = 0 \\
   \Rightarrow k = 2 \\
 $
Putting $k = 2$ in $R(3 - k,k - 4,6k - 5)$ we have
$R(3 - k,k - 4,6k - 5) = R(3 - 2,2 - 4,6 \times 2 - 5)$
Therefore, we get the coordinate as
$R(1, - 2,7)$
Thus, the coordinate of the required point is $(1, - 2,7)$ .

Note: In this type of question students often find it difficult to proceed with the first step. Try to figure out the hints from the question like “crosses the plane” and all those things similar to this.