Question

Find the equation of the line passing through the point \[(-1,3,-2)\] and perpendicular to the line
\[\dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z}{3}\] and \[\dfrac{x+2}{-3}=\dfrac{y-1}{2}=\dfrac{z+1}{5}\].

Answer 1

Hint: First we will calculate the slope of the axis of that unknown line which is passing through point \[(-1,3,-2)\]. It is given that the slope of line is perpendicular to both given lines. So we will do the cross product of slopes of both given lines and now we have slope and point from which it is passing is given now we can calculate equation using formula \[\dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}\] where \[a,b,c\] are slopes and \[{{x}_{1}},{{y}_{1}},{{z}_{1}}\] are the point from which it passes .

Complete step by step answer:
We have to find a line which passes through a given point \[(-1,3,-2)\] and it is perpendicular to two lines \[\dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z}{3}\] and \[\dfrac{x+2}{-3}=\dfrac{y-1}{2}=\dfrac{z+1}{5}\]
So first we have to find the slope of that unknown line, by looking at given lines we can say that there slopes are \[(1,2,3)\]and \[(-3,2,5)\] which can be written in vector form as \[i+2j+3k\] and \[-3i+2j+5k\] now we have to do cross product of these two slopes
Using formal of cross product: \[(ai+bj+ck)\times (di+ej+fk)=\left( \begin{matrix}
   i & j & k \\
   a & b & c \\
   d & e & f \\
\end{matrix} \right)\]
\[\to (i+2j+3k)\times (-3i+2j+5k)\]
Which is equal to \[\left( \begin{matrix}
   i & j & k \\
   1 & 2 & 3 \\
   -3 & 2 & 5 \\
\end{matrix} \right)\]
Which on solving equals to \[4i-14j+8k\]
So \[4i-14j+8k\] this is our slope of unknown line and the line passes through point \[(-1,3,-2)\]
Now we have slope as well as point passing so we can calculate equation of line by using
\[\dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}\] where a,b,c are slopes and \[{{x}_{1}},{{y}_{1}},{{z}_{1}}\] are the point from which it passes
Slope is \[4i-14j+8k=ai+bj+ck\] so we can substitute \[a=4,b=-14,c=8\] and point \[{{x}_{1}},{{y}_{1}},{{z}_{1}}\]
By \[(-1,3,-2)\] so our unknown line equation is \[\dfrac{x-(-1)}{4}=\dfrac{y-3}{-14}=\dfrac{z-(-2)}{8}\]
Hence the equation of unknown line is \[\dfrac{x+1}{4}=\dfrac{y-3}{-14}=\dfrac{z+2}{8}\]

Note:
Most of the time we make mistakes while opening the matrix. For example we forget to take minus signs in front of the j vector. if it is given that unknown line passes through two given points then also, we can find its equation for example, two given points are
\[(1,2,3)\] and \[(4,5,6)\] now slope of this line will be \[(4-1,5-2,6-3)=3i+3j+3k\]
And it passes through \[(1,2,3)\]hence line equation will be \[\dfrac{x-1}{3}=\dfrac{y-2}{3}=\dfrac{z-3}{3}\].