Answer
Verified
414.9k+ views
Hint: In this question, we are given coordinates of $A$and coordinates of a point $C\left( -2,3,5 \right)$ which divides the line $AB$ in the ratio $1:3$. We have to find the coordinates of point $B$. For this, we will first suppose the coordinates of $B$ and then apply section formula on the line joining $AB$ which is cut by a point in the ratio $1:3$. Section formula on any line with $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ as coordinates of end point cut by a point $\left( {{x}_{3}},{{y}_{3}},{{z}_{3}} \right)$ in the ratio $m:n$is given by –
$\left( {{x}_{3}},{{y}_{3}},{{z}_{3}} \right)=\left( \dfrac{n{{x}_{1}}+m{{x}_{2}}}{m+n},\dfrac{n{{y}_{1}}+m{{y}_{2}}}{m+n},\dfrac{n{{z}_{1}}+m{{z}_{2}}}{m+n} \right)$
Complete step-by-step solution
We are given coordinates of point $A$ as $\left( 1,3,4 \right)$. Let us suppose coordinates of $B$ as $\left( x,y,z \right)$. We are also given coordinates of points which cut the line as $\left( -2,3,5 \right)$. Ratio is given as $1:3$. Hence, by comparing with section formula,
$\begin{align}
& m:n=1:3 \\
& \left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)=\left( 1,3,4 \right) \\
& \left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)=\left( x,y,z \right) \\
& \left( {{x}_{3}},{{y}_{3}},{{z}_{3}} \right)=\left( -2,3,5 \right) \\
\end{align}$
Putting in the formula, we get –
$\left( -2,3,5 \right)=\left( \dfrac{3\left( 1 \right)+1\left( x \right)}{3+1},\dfrac{3\left( 3 \right)+1\left( y \right)}{3+1},\dfrac{3\left( 4 \right)+1\left( z \right)}{3+1} \right)$
Comparing and separating the terms, we get –
$-2=\dfrac{3\left( 1 \right)+1\left( x \right)}{3+1},3=\dfrac{3\left( 3 \right)+1\left( y \right)}{3+1},5=\dfrac{3\left( 4 \right)+1\left( z \right)}{3+1}$
Solving for value of $x$, we get –
$-2=\dfrac{3+x}{4}$
Cross multiplying, we get –
$\begin{align}
& -8=3+x \\
& x=-11……………......(1) \\
\end{align}$
Solving for value of $y$, we get –
$3=\dfrac{9+y}{4}$
Cross multiplying, we get –
$\begin{align}
& 12=9+y \\
& y=3……….....(2) \\
\end{align}$
Solving for value of $z$, we get –
$5=\dfrac{12+z}{4}$
Cross multiplying, we get –
$\begin{align}
& 20=12+z \\
& z=8 …………....(3) \\
\end{align}$
From equation (1), (2) and (3), we get –
$\left( x,y,z \right)=\left( -11,3,8 \right)$
As we have supposed coordinates of $B$ as $\left( x,y,z \right)$. Therefore, the coordinates of $B$ are $\left( -11,3,8 \right)$.
Hence, option (a) is the correct answer.
Note: Take care in the order of $A$ and $B$. If line $BA$ would have been given, the answer would have changed completely because ratio changes, which changes the solution completely. Also try to draw a diagram so that we can know if the point divides the line internally or externally. For externally, the formula changes accordingly and do not get confused in the value of $m$ and $n$, and also in values of $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and \[\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)\].
$\left( {{x}_{3}},{{y}_{3}},{{z}_{3}} \right)=\left( \dfrac{n{{x}_{1}}+m{{x}_{2}}}{m+n},\dfrac{n{{y}_{1}}+m{{y}_{2}}}{m+n},\dfrac{n{{z}_{1}}+m{{z}_{2}}}{m+n} \right)$
Complete step-by-step solution
We are given coordinates of point $A$ as $\left( 1,3,4 \right)$. Let us suppose coordinates of $B$ as $\left( x,y,z \right)$. We are also given coordinates of points which cut the line as $\left( -2,3,5 \right)$. Ratio is given as $1:3$. Hence, by comparing with section formula,
$\begin{align}
& m:n=1:3 \\
& \left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)=\left( 1,3,4 \right) \\
& \left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)=\left( x,y,z \right) \\
& \left( {{x}_{3}},{{y}_{3}},{{z}_{3}} \right)=\left( -2,3,5 \right) \\
\end{align}$
Putting in the formula, we get –
$\left( -2,3,5 \right)=\left( \dfrac{3\left( 1 \right)+1\left( x \right)}{3+1},\dfrac{3\left( 3 \right)+1\left( y \right)}{3+1},\dfrac{3\left( 4 \right)+1\left( z \right)}{3+1} \right)$
Comparing and separating the terms, we get –
$-2=\dfrac{3\left( 1 \right)+1\left( x \right)}{3+1},3=\dfrac{3\left( 3 \right)+1\left( y \right)}{3+1},5=\dfrac{3\left( 4 \right)+1\left( z \right)}{3+1}$
Solving for value of $x$, we get –
$-2=\dfrac{3+x}{4}$
Cross multiplying, we get –
$\begin{align}
& -8=3+x \\
& x=-11……………......(1) \\
\end{align}$
Solving for value of $y$, we get –
$3=\dfrac{9+y}{4}$
Cross multiplying, we get –
$\begin{align}
& 12=9+y \\
& y=3……….....(2) \\
\end{align}$
Solving for value of $z$, we get –
$5=\dfrac{12+z}{4}$
Cross multiplying, we get –
$\begin{align}
& 20=12+z \\
& z=8 …………....(3) \\
\end{align}$
From equation (1), (2) and (3), we get –
$\left( x,y,z \right)=\left( -11,3,8 \right)$
As we have supposed coordinates of $B$ as $\left( x,y,z \right)$. Therefore, the coordinates of $B$ are $\left( -11,3,8 \right)$.
Hence, option (a) is the correct answer.
Note: Take care in the order of $A$ and $B$. If line $BA$ would have been given, the answer would have changed completely because ratio changes, which changes the solution completely. Also try to draw a diagram so that we can know if the point divides the line internally or externally. For externally, the formula changes accordingly and do not get confused in the value of $m$ and $n$, and also in values of $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and \[\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)\].
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Why Are Noble Gases NonReactive class 11 chemistry CBSE
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
At which age domestication of animals started A Neolithic class 11 social science CBSE
Which are the Top 10 Largest Countries of the World?
Give 10 examples for herbs , shrubs , climbers , creepers
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
Write a letter to the principal requesting him to grant class 10 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE