Question

The x=coordinate of a point on the line joining the points P(2,2,1) and Q(5,1,-2) is 4. Find its z-coordinate.

Answer 1

This question is based in coordinate geometry. You are given two points in space and the coordinate points on the line joining the given points. Find the z-coordinate of that point. You need to know the equation of a line joining two parts in space to solve this problem.

Step wise Solution:
Given data: The points are given as P(2,2,1) and Q(5,1,-2) also, the x-coordinate of a point on the line joining P and Q is given as 4.
We need to find the z-coordinate of that point that lies on the line joining P and Q, whose x-coordinate is 4.
To compute the equation of the line joining P and Q.
We know that, equation of a straight line joining two points $A({x_1},{y_1},{z_1})\,\,and\,\,B({x_2},{y_2},{z_2})$ is given by $\dfrac{{x - {x_1}}}{{{x_2} - {x_1}}} = \dfrac{{y - {y_1}}}{{{y_2} - {y_1}}} = \dfrac{{z - {z_1}}}{{{z_2} - {z_1}}}$
For, the points P(2,2,1) and Q(5,1,-2) , the equation of the line joining the points P and Q is given by,

$$\dfrac{{x - 2}}{{5 - 2}} = \dfrac{{y - 2}}{{1 - 2}} = \dfrac{{z - 1}}{{ - 2 - 1}}\\\ \Rightarrow \dfrac{{x - 2}}{3} = \dfrac{{y - 2}}{{ - 1}} = \dfrac{{z - 1}}{{ - 3}}$$

Is the required equation of the line joining points P(2,2,1) and (5,1,-2)
Now, To find out the z-coordinate of a point lying on the line joining the points P and Q, where x-coordinate is 4.
Suppose,

\dfrac{{x - 2}}{3} = \dfrac{{y - 2}}{{ - 1}} = \dfrac{{z - 1}}{{ - 3}} = r\\\ $$ which $$ \,\,gives,\\\\$$ $$\Rightarrow \dfrac{{x - 2}}{3} = r\\\ $$ $$\Rightarrow x - 2 = 3r\\\ $$ $$\Rightarrow x = 2 + 3r .....(i) \\\ $$ And, $$ \Rightarrow \dfrac{{y - 2}}{{ - 1}} = r\\\ $$ $$ \Rightarrow y - 2 = - r\\\ $$ $$ \Rightarrow y = 2 - r .....(ii) \\\ $$ Also, $$ \Rightarrow \dfrac{{z - 1}}{{ - 3}} = r\\\ $$ $$ \Rightarrow z - 1 = - 3r\\\ $$ $$ \Rightarrow z = 1 - 3r .....(iii) \\\ $$ Let us consider a point T on the line joining the two points P(2,2,1) and Q(5,1,-2). From the equation(i), (ii),(iii), we have $$T(x,y,z) = T(2 + 3r,2 - r,1 - 3r)$$ As we are given in the question itself that the x-coordinate of the point is 4. Which means,

x = 4\\
\Rightarrow 2 + 3r = 4\\
\Rightarrow 3r = 4 - 2\\
\Rightarrow r = \dfrac{2}{3}

$$We need to find the z-coordinate of T(x,y,z). So lets put the value of r in equation(iii) Equation (iii)$$

z = 1 - 3r\\
\Rightarrow z = 1 - {3} \times \dfrac{2}{{{3}}}\\
\Rightarrow z = 1 - 2\\
\Rightarrow z = - 1 $$

Hence, the z-coordinate of the pint is -1.

Note: You can also solve this question by using section formula. All you need to do is to consider a point T divides the line joining the points P(2,2,1) and Q(5,1,-2) in the ratio $\lambda :1$ , then find the value of $\lambda$ and thereafter you can compute the z-coordinate of the point.