Answer
Verified
455.4k+ views
Hint: First, before proceeding for this, we must assume a factor $\lambda $ which is called a proportionality factor of the given line equation. Then, by equation the given equation of line to the factor $\lambda $, we get all the points. Then, we know the condition that if the angle between the lines is ${{90}^{\circ }}$, then their dot product of direction ratios are zero. Then, by using this condition we get the value of $\lambda $and by using that we get the points. Then, by using the distance formula as $d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{z}_{2}}-{{z}_{1}} \right)}^{2}}}$, we get the desired answer.
Complete step-by-step solution:
In this question, we are supposed to find the length of the perpendicular from P(2, -3, 1) to the line $\dfrac{x-1}{2}=\dfrac{y-3}{3}=\dfrac{z+2}{-1}$.
So, before proceeding for this, we must assume a factor $\lambda $which is called a proportionality factor of the given line equation.
Now, by equation the given equation of line to the factor $\lambda $, we get:
$\dfrac{x-1}{2}=\dfrac{y-3}{3}=\dfrac{z+2}{-1}=\lambda $
Then, we get the points as:
$\begin{align}
& \dfrac{x-1}{2}=\lambda \\
& \Rightarrow x-1=2\lambda \\
& \Rightarrow x=2\lambda +1 \\
\end{align}$
Similarly, we get for the other two conditions, the points are as:
$\left( 2\lambda +1,\text{ }3\lambda +3,\text{ }-\lambda -2 \right)$
Then, we are given with the point P(2, -3, 1) which acts like direction ratio of line which is perpendicular to the given line as:
$\begin{align}
& \left( 2\lambda +1-2,\text{ }3\lambda +3+3,\text{ }-\lambda -2-1 \right) \\
& \Rightarrow \left( 2\lambda -1,\text{ }3\lambda +6,\text{ }-\lambda -3 \right) \\
\end{align}$
So, from the above equation, we get the direction ratios of the line by the coefficients of the factor $\lambda $ as 2, 3 and -1.
Now, we got the expression for the points of the line which is perpendicular to the given line.
So, we know the condition that if the angle between the lines is ${{90}^{\circ }}$, then their dot product of direction ratios are zero.
Then, by using this condition, we get:
$\begin{align}
& \left( 2\lambda -1 \right)\centerdot 2+\left( 3\lambda +6 \right)\centerdot 3+\left( -\lambda -3 \right)\centerdot \left( -1 \right)=0 \\
& \Rightarrow 4\lambda -2+9\lambda +18+\lambda +3=0 \\
& \Rightarrow 14\lambda +19=0 \\
& \Rightarrow 14\lambda =-19 \\
& \Rightarrow \lambda =\dfrac{-19}{14} \\
\end{align}$
Now, we will substitute the value of $\lambda =\dfrac{-19}{14}$ in the expression found as $\left( 2\lambda +1,\text{ }3\lambda +3,\text{ }-\lambda -2 \right)$ to get the points of the line as:
$\begin{align}
& 2\left( \dfrac{-19}{14} \right)+1,\text{ 3}\left( \dfrac{-19}{14} \right)+3,\text{ }-\left( \dfrac{-19}{14} \right)-2 \\
& \Rightarrow \dfrac{-38+14}{14},\text{ }\dfrac{-57+42}{14},\text{ }\dfrac{19-28}{14} \\
& \Rightarrow \dfrac{-24}{14},\text{ }\dfrac{-15}{14},\text{ }\dfrac{-9}{14} \\
\end{align}$
So, we get the points of the line as $\left( \dfrac{-24}{14},\text{ }\dfrac{-15}{14},\text{ }\dfrac{-9}{14} \right)$ which is perpendicular to points (1, 3, -2).
Now, we will use the distance formula to calculate the distance(d) between these points $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$and $\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ as:
$d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{z}_{2}}-{{z}_{1}} \right)}^{2}}}$
So, by using the formula for the points found above, we get:
$\begin{align}
& d=\sqrt{{{\left( 2+\dfrac{24}{14} \right)}^{2}}+{{\left( -3+\dfrac{15}{14} \right)}^{2}}+{{\left( 1+\dfrac{9}{14} \right)}^{2}}} \\
& \Rightarrow d=\sqrt{{{\left( \dfrac{52}{14} \right)}^{2}}+{{\left( \dfrac{-27}{14} \right)}^{2}}+{{\left( \dfrac{23}{14} \right)}^{2}}} \\
& \Rightarrow d=\sqrt{\dfrac{2704}{196}+\dfrac{729}{196}+\dfrac{529}{196}} \\
& \Rightarrow d=\sqrt{\dfrac{3962}{196}} \\
& \Rightarrow d=\sqrt{\dfrac{531}{14}} \\
\end{align}$
So, the distance between the points is $\sqrt{\dfrac{531}{14}}$.
Hence, option (b) is correct.
Note: Now, to solve this type of questions we must know the angle condition of the direction ratios which is used above in the question as both the lines are perpendicular so the dot product is zero is given by the condition that:
$\cos \theta =\dfrac{\left| {{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}} \right|}{\sqrt{{{a}_{1}}^{2}+{{b}_{1}}^{2}+{{c}_{1}}^{2}}\sqrt{{{a}_{2}}^{2}+{{b}_{2}}^{2}+{{c}_{2}}^{2}}}$
Where $\left( {{a}_{1}},{{b}_{1}},{{c}_{1}} \right)$ and $\left( {{a}_{2}},{{b}_{2}},{{c}_{2}} \right)$ are the points of the line.
Complete step-by-step solution:
In this question, we are supposed to find the length of the perpendicular from P(2, -3, 1) to the line $\dfrac{x-1}{2}=\dfrac{y-3}{3}=\dfrac{z+2}{-1}$.
So, before proceeding for this, we must assume a factor $\lambda $which is called a proportionality factor of the given line equation.
Now, by equation the given equation of line to the factor $\lambda $, we get:
$\dfrac{x-1}{2}=\dfrac{y-3}{3}=\dfrac{z+2}{-1}=\lambda $
Then, we get the points as:
$\begin{align}
& \dfrac{x-1}{2}=\lambda \\
& \Rightarrow x-1=2\lambda \\
& \Rightarrow x=2\lambda +1 \\
\end{align}$
Similarly, we get for the other two conditions, the points are as:
$\left( 2\lambda +1,\text{ }3\lambda +3,\text{ }-\lambda -2 \right)$
Then, we are given with the point P(2, -3, 1) which acts like direction ratio of line which is perpendicular to the given line as:
$\begin{align}
& \left( 2\lambda +1-2,\text{ }3\lambda +3+3,\text{ }-\lambda -2-1 \right) \\
& \Rightarrow \left( 2\lambda -1,\text{ }3\lambda +6,\text{ }-\lambda -3 \right) \\
\end{align}$
So, from the above equation, we get the direction ratios of the line by the coefficients of the factor $\lambda $ as 2, 3 and -1.
Now, we got the expression for the points of the line which is perpendicular to the given line.
So, we know the condition that if the angle between the lines is ${{90}^{\circ }}$, then their dot product of direction ratios are zero.
Then, by using this condition, we get:
$\begin{align}
& \left( 2\lambda -1 \right)\centerdot 2+\left( 3\lambda +6 \right)\centerdot 3+\left( -\lambda -3 \right)\centerdot \left( -1 \right)=0 \\
& \Rightarrow 4\lambda -2+9\lambda +18+\lambda +3=0 \\
& \Rightarrow 14\lambda +19=0 \\
& \Rightarrow 14\lambda =-19 \\
& \Rightarrow \lambda =\dfrac{-19}{14} \\
\end{align}$
Now, we will substitute the value of $\lambda =\dfrac{-19}{14}$ in the expression found as $\left( 2\lambda +1,\text{ }3\lambda +3,\text{ }-\lambda -2 \right)$ to get the points of the line as:
$\begin{align}
& 2\left( \dfrac{-19}{14} \right)+1,\text{ 3}\left( \dfrac{-19}{14} \right)+3,\text{ }-\left( \dfrac{-19}{14} \right)-2 \\
& \Rightarrow \dfrac{-38+14}{14},\text{ }\dfrac{-57+42}{14},\text{ }\dfrac{19-28}{14} \\
& \Rightarrow \dfrac{-24}{14},\text{ }\dfrac{-15}{14},\text{ }\dfrac{-9}{14} \\
\end{align}$
So, we get the points of the line as $\left( \dfrac{-24}{14},\text{ }\dfrac{-15}{14},\text{ }\dfrac{-9}{14} \right)$ which is perpendicular to points (1, 3, -2).
Now, we will use the distance formula to calculate the distance(d) between these points $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$and $\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ as:
$d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{z}_{2}}-{{z}_{1}} \right)}^{2}}}$
So, by using the formula for the points found above, we get:
$\begin{align}
& d=\sqrt{{{\left( 2+\dfrac{24}{14} \right)}^{2}}+{{\left( -3+\dfrac{15}{14} \right)}^{2}}+{{\left( 1+\dfrac{9}{14} \right)}^{2}}} \\
& \Rightarrow d=\sqrt{{{\left( \dfrac{52}{14} \right)}^{2}}+{{\left( \dfrac{-27}{14} \right)}^{2}}+{{\left( \dfrac{23}{14} \right)}^{2}}} \\
& \Rightarrow d=\sqrt{\dfrac{2704}{196}+\dfrac{729}{196}+\dfrac{529}{196}} \\
& \Rightarrow d=\sqrt{\dfrac{3962}{196}} \\
& \Rightarrow d=\sqrt{\dfrac{531}{14}} \\
\end{align}$
So, the distance between the points is $\sqrt{\dfrac{531}{14}}$.
Hence, option (b) is correct.
Note: Now, to solve this type of questions we must know the angle condition of the direction ratios which is used above in the question as both the lines are perpendicular so the dot product is zero is given by the condition that:
$\cos \theta =\dfrac{\left| {{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}} \right|}{\sqrt{{{a}_{1}}^{2}+{{b}_{1}}^{2}+{{c}_{1}}^{2}}\sqrt{{{a}_{2}}^{2}+{{b}_{2}}^{2}+{{c}_{2}}^{2}}}$
Where $\left( {{a}_{1}},{{b}_{1}},{{c}_{1}} \right)$ and $\left( {{a}_{2}},{{b}_{2}},{{c}_{2}} \right)$ are the points of the line.
Recently Updated Pages
what is the correct chronological order of the following class 10 social science CBSE
Which of the following was not the actual cause for class 10 social science CBSE
Which of the following statements is not correct A class 10 social science CBSE
Which of the following leaders was not present in the class 10 social science CBSE
Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE
Which one of the following places is not covered by class 10 social science CBSE
Trending doubts
Which are the Top 10 Largest Countries of the World?
The states of India which do not have an International class 10 social science CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you graph the function fx 4x class 9 maths CBSE
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE
Difference Between Plant Cell and Animal Cell
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE
Name the three parallel ranges of the Himalayas Describe class 9 social science CBSE