Question

How do you find an equation for the plane that contains the line with parametric equations $l=\left( \left( 8-7t \right),\left( -5-2t \right),\left( 5-t \right) \right)$ and is parallel to the line with parametric equations $x=3+t$, $y=-7+9t$, $z=8-6t$?

Answer 1

Consider the parametric equations of the line that is parallel to the plane and find the values of t in terms of x, y and z one – by – one. Reduce it to the form: $\dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}$, where a, b and c are called the direction ratios. Now, assume the equation of the required plane as- $\dfrac{x-p}{a}=\dfrac{y-q}{b}=\dfrac{z-r}{c}$, where p, q and r denotes the point from which the plane passes. To determine p, q and r substitute t = 0 in the parametric equation of line l.

Complete step by step solution:
Here, we have been provided with two lines and we are asked to determine the equation of a plane that is parallel to one of the lines and contains the other line.
Now, it is said that the plane is parallel to the line with parametric equations $x=3+t$, $y=-7+9t$, $z=8-6t$. That means the direction ratios of the plane and line must be the same. For a line or plane of the form: - $\dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}$ we have a, b and c as its direction ratios. So, converting the given line into the above form, we get,
$\Rightarrow \dfrac{x-3}{1}=\dfrac{y+7}{9}=\dfrac{z-8}{-6}$
Therefore, the direction ratios of the line and hence the plane is given as (1, 9, -6). So, the equation of the plane can be assumed as: - $\dfrac{x-p}{1}=\dfrac{y-q}{9}=\dfrac{z-r}{-6}$, where p, q and r denotes the point from which the plane passes. We need to determine these values.
Now, since the plane contains the line $l=\left( 8-7t,-5-2t,5-t \right)$ as given in the question, so each and every point on this line must also lie on the plane. We need to determine the coordinate of any one such point. Substituting the value of t equal to 0, we get,
$\Rightarrow l=\left( 8,-5,5 \right)$
Therefore, the above point lies on the line and hence the plane passes through this point.
$\Rightarrow \left( p,q,r \right)=\left( 8,-5,5 \right)$
So, substituting the values of p, q and r in the assumed equation of the plane, we get,
$\Rightarrow \dfrac{x-8}{1}=\dfrac{y+5}{9}=\dfrac{z-5}{-6}$
Hence, the above equation is our answer.

Note:
One may note that whenever two lines or a line and a plane are parallel then their direction ratio or you can say direction cosines are equal. In the case if they are perpendicular then the product of their direction ratios equals 0. You must remember the two conditions. Note that you can also substitute different values of ‘t’ to get the point (p, q, r). For example: - for t = 1 we have the point as l = (1, -7, 4). This point will also lie on the plane.