Question

How do you find a vector equation and parametric equations in $t$ for the line through the point and perpendicular to the given plane. ($PO$ corresponds to $t=0$ .) $PO=\left( 5,0,8 \right)x+2y+z=9$ ?

Answer 1

Here we need to find the equation of a vector and parametric equations in $t$ for the line through the point and perpendicular to the given plane. ($PO$ corresponds to $t=0$ .) $PO=\left( 5,0,8 \right)x+2y+z=9$. we know that the vector equation of line passing through a point $A$ (having position vector $\vec{a}$ ) and having direction vector along $\vec{l}$ is given by $\vec{r}=\vec{a}+t.\vec{l}$ . Its Cartesian equation for $A=\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and $\vec{l}=\left( g,h,f \right)$ is given by $\dfrac{x-{{x}_{1}}}{g}=\dfrac{y-{{y}_{1}}}{f}=\dfrac{z-{{z}_{1}}}{h}$ .
Complete step by step solution:
Now considering from the question we have been asked to find the equation of a vector and parametric equations in $t$ for the line through the point and perpendicular to the given plane. ($PO$ corresponds to $t=0$ .) $PO=\left( 5,0,8 \right)x+2y+z=9$.
Let us assume $L$ is the required line.
We can say that $L$ is perpendicular to the given plane $P:x+2y+z=9$ .
So now if we assume that the direction vector of $L$ is $\vec{l}$ . We know that this vector will be parallel to the normal of $P$ . So, if we assume the normal of $P$ is $\vec{n}$ then it will have the value $<1,2,1>$ .
Now from the basic concepts we know that the vector equation of line passing through a point $A$ (having position vector $\vec{a}$ ) and having direction vector along $\vec{l}$ is given by $\vec{r}=\vec{a}+t.\vec{l}$ . Its Cartesian equation for $A=\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and $\vec{l}=\left( g,h,f \right)$ is given by $\dfrac{x-{{x}_{1}}}{g}=\dfrac{y-{{y}_{1}}}{f}=\dfrac{z-{{z}_{1}}}{h}$.
Hence in our case, the required line equation will be $<5,0,8>+t.<1,2,1>$
The Cartesian equation will be given as $\dfrac{x-5}{1}=\dfrac{y-0}{2}=\dfrac{z-8}{1}$ .
Therefore we can conclude that the answer for this question is given as $x-5=\dfrac{y}{2}=z-8$ .

Note: During the process of answering questions of this type we should be sure with the concepts that we are going to apply. This is a very confusing question and mistakes are very much possible in this type of question. To avoid and reduce mistakes we need to practice more and more questions. For example we can interchange the normal and point and write the equation as $<1,2,1>+t.<5,0,8>$ then we will have the equation as $\dfrac{x-1}{5}=\dfrac{y-2}{0}=\dfrac{z-1}{8}$ which is a wrong answer.