
Find the ratio in which the XY - plane divides AB if coordinates of A is (1,2,3) and B is (-3,4,-5). Also find the positive vector of the point of division.
Answer
619.5k+ views
Hint: To solve this problem, we would need the basic concepts of three dimensional geometry and vectors. We further, use the formula for a point to divide line segment that is $\left( \dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n},\dfrac{m{{z}_{2}}+n{{z}_{1}}}{m+n} \right)$, if point divides the line segment in ratio m:n.
Complete step by step solution:
We first find the line segment AB given by A (1,2,3) and B (−3,4,−5). Now, we are to find the ratio in which the XY – plane divides AB. Now, we assume that the XZ−plane divides line AB at point C (x,y,z) in the ratio k:1. Now, the formula for a point C (x,y,z) to divide line segment joining point A ( ${{x}_{1}},{{y}_{1}},{{z}_{1}}$ ) and B( ${{x}_{2}},{{y}_{2}},{{z}_{2}}$ ) in the ratio m:n is given by-
= $\left( \dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n},\dfrac{m{{z}_{2}}+n{{z}_{1}}}{m+n} \right)$
Here, in this case, we are given that the ratio as k:1. Thus, we will use the above formula with m = k and n = 1. Now, we have,
A ( ${{x}_{1}},{{y}_{1}},{{z}_{1}}$ ) = (1,2,3)
B ( ${{x}_{2}},{{y}_{2}},{{z}_{2}}$ ) = (−3,4,−5)
Now, using the above formula, we can find the coordinate of C (x,y,z) as -
C (x, y, z) = $\left( \dfrac{k{{x}_{2}}+{{x}_{1}}}{k+1},\dfrac{k{{y}_{2}}+{{y}_{1}}}{k+1},\dfrac{k{{z}_{2}}+{{z}_{1}}}{k+1} \right)$
(Since, m = k and n = 1)
Now, putting values of points A ( ${{x}_{1}},{{y}_{1}},{{z}_{1}}$ ) and B( ${{x}_{2}},{{y}_{2}},{{z}_{2}}$ ) as (1, 2, 3) and (-3, 4, -5). Thus, we have,
= $\left( \dfrac{k(-3)+1}{k+1},\dfrac{k(4)+2}{k+1},\dfrac{k(-5)+3}{k+1} \right)$ -- (1)
Since point C (x,y,z) lies on the XY – plane, the z coordinate of C will be zero. Thus, we have,
-5k + 3 = 0
k = $\dfrac{3}{5}$
We can substitute the value of k in (1), we get,
= $\left( \dfrac{-0.8}{1.6},\dfrac{4.4}{1.6},0 \right)$
= (-0.5, 2.75, 0) -- (2)
Thus, the required ratio is k:1, that is 0.6 : 1. Simplifying the ratio, we get,
0.6 : 1 = 3 : 5. Thus, the required ratio is 3 : 5.
We get the positive vector of the point of division from (2). Thus, the vector is given by -0.5i + 2.75j. Here, i denotes unit vector to X-axis and j denotes unit vector to Y-axis.
Note: Generally, while solving the problems related to finding points dividing a line segment in a particular ratio, we always try to use the formula by inserting ratio as k : 1, instead of m:n. This way, we only have to deal with one variable, which makes it easier to deal with the obtained equations.
Complete step by step solution:
We first find the line segment AB given by A (1,2,3) and B (−3,4,−5). Now, we are to find the ratio in which the XY – plane divides AB. Now, we assume that the XZ−plane divides line AB at point C (x,y,z) in the ratio k:1. Now, the formula for a point C (x,y,z) to divide line segment joining point A ( ${{x}_{1}},{{y}_{1}},{{z}_{1}}$ ) and B( ${{x}_{2}},{{y}_{2}},{{z}_{2}}$ ) in the ratio m:n is given by-
= $\left( \dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n},\dfrac{m{{z}_{2}}+n{{z}_{1}}}{m+n} \right)$
Here, in this case, we are given that the ratio as k:1. Thus, we will use the above formula with m = k and n = 1. Now, we have,
A ( ${{x}_{1}},{{y}_{1}},{{z}_{1}}$ ) = (1,2,3)
B ( ${{x}_{2}},{{y}_{2}},{{z}_{2}}$ ) = (−3,4,−5)
Now, using the above formula, we can find the coordinate of C (x,y,z) as -
C (x, y, z) = $\left( \dfrac{k{{x}_{2}}+{{x}_{1}}}{k+1},\dfrac{k{{y}_{2}}+{{y}_{1}}}{k+1},\dfrac{k{{z}_{2}}+{{z}_{1}}}{k+1} \right)$
(Since, m = k and n = 1)
Now, putting values of points A ( ${{x}_{1}},{{y}_{1}},{{z}_{1}}$ ) and B( ${{x}_{2}},{{y}_{2}},{{z}_{2}}$ ) as (1, 2, 3) and (-3, 4, -5). Thus, we have,
= $\left( \dfrac{k(-3)+1}{k+1},\dfrac{k(4)+2}{k+1},\dfrac{k(-5)+3}{k+1} \right)$ -- (1)
Since point C (x,y,z) lies on the XY – plane, the z coordinate of C will be zero. Thus, we have,
-5k + 3 = 0
k = $\dfrac{3}{5}$
We can substitute the value of k in (1), we get,
= $\left( \dfrac{-0.8}{1.6},\dfrac{4.4}{1.6},0 \right)$
= (-0.5, 2.75, 0) -- (2)
Thus, the required ratio is k:1, that is 0.6 : 1. Simplifying the ratio, we get,
0.6 : 1 = 3 : 5. Thus, the required ratio is 3 : 5.
We get the positive vector of the point of division from (2). Thus, the vector is given by -0.5i + 2.75j. Here, i denotes unit vector to X-axis and j denotes unit vector to Y-axis.
Note: Generally, while solving the problems related to finding points dividing a line segment in a particular ratio, we always try to use the formula by inserting ratio as k : 1, instead of m:n. This way, we only have to deal with one variable, which makes it easier to deal with the obtained equations.
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

