Question

Find the vector equation of the line passing through the point \[2i+3j+\hat{k}\] and parallel to the vector \[4i-2j+3k\].

Answer 1

Hint: We know that if a line is passing through the point \[\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)\] and parallel to the vector \[\left( ai+bj+ck \right)\] is given as:
\[\dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}\]
So by using this formula, we will find the equation of the required line.

Complete step-by-step answer:
We have been asked to find the equation of a line passing through the point \[2i+3j+\hat{k}\] and parallel to the vector \[4i-2j+3k\].
So the line is passing through the point (2, 3, 1) in Cartesian form
We know that if a line is passing through the point \[\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)\] and parallel to the vector \[\left( ai+bj+ck \right)\] is given as:
\[\dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}\]
Hence the equation of the line passing through (2, 3, 1) and parallel to the vector \[4i-2j+3k\] is given as:
\[\begin{align}
  & \dfrac{x-2}{4}=\dfrac{y-3}{-2}=\dfrac{z-1}{3} \\
 & \dfrac{x-2}{4}=\dfrac{-y+3}{2}=\dfrac{z-1}{3} \\
\end{align}\]
Also, we know that a line passing through the point \[x\hat{i}+y\hat{j}+z\hat{k}\] and parallel to vector \[a\hat{i}+b\hat{j}+c\hat{k}\] is represented in vector form as follows:
\[r\left( t \right)=\left( x,y,z \right)+t\left( a,b,c \right)\]
Therefore the required equation is \[r\left( t \right)=\left( 2,3,1 \right)+t\left( 4,-2,3 \right)\].

Note: Be careful while finding the equation of line and also take care of the sign of the coordinate during substituting its values in formula to find the equation of line. Remember that the vector equation of a line passing through the point \[x\hat{i}+y\hat{j}+z\hat{k}\] and parallel to the vector \[a\hat{i}+b\hat{j}+c\hat{k}\] is given by \[r\left( t \right)=\left( x,y,z \right)+t\left( a,b,c \right)\].