
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
586.2k+ views
Hint: 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.
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.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

What are the major means of transport Explain each class 12 social science CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE

Why cannot DNA pass through cell membranes class 12 biology CBSE

Differentiate between insitu conservation and exsitu class 12 biology CBSE

Draw a neat and well labeled diagram of TS of ovary class 12 biology CBSE

