91Ó°ÊÓ

To express the answer as a vector parametrization first, we have to find the direction vector for the line \(\overrightarrow{P Q}\).

So,

\(\overrightarrow{P Q}=\left(4-1,5-6\right)\)

\(\overrightarrow{P Q}=\left(3,-1\right)\).

Now, the formula to find the line equation is \(r(t)=P_{0}+td\).

So,

\(r(t)=\left ( 1,6 \right )+t\left ( 3,-1 \right )\)

\(r(t)=\left ( 1+3t,6-t \right )\)

Thus, the equation of a line in the form of vector parametrization is \(r(t)=\left ( 1+3t,6-t \right )\).

As we get from part (a) that the equation of a line in the form of vector parametrization is \(r(t)=\left ( 1+3t,6-t \right )\).

Now, the vector function r(t) in 2-dimesional represent as \(r(t)=\left ( x\left ( t \right ),y\left ( t \right ) \right )\).

So, the parametric equations are \(x\left ( t \right )=1+3t,y\left ( t \right )=6-t\).

To express the equation of a line in a symmetric form. We have to eliminate the parameter t from the parametric equations of a line.

Now, the parametric equations are \(x\left ( t \right )=1+3t,y\left ( t \right )=6-t\).

Let's take

\(x\left ( t \right )=1+3t\)

\(x=1+3t\)

\(x-1=3t\)

\(\frac{x-1}{3}=t\) ........(i)

Now, take

\(y\left ( t \right )=6-t\)

\(y=6-t\)

\(y-6=-t\)

\(\left ( -1 \right )\left ( y-6 \right )=\left ( -1 \right )\left ( -t \right )\)

\(-y+6=t\)\) ........(ii)

By equating the equations (i) and (ii) we get

\(\frac{x-1}{3}=t,-y+6=t\).

Therefore, \(\frac{x-1}{3}=-y+6=t\).

Hence, the equation in symmetric form is \(\frac{x-1}{3}=-y+6\).