Question

Equation of the projection of the line $8x-y-7z=8$ , $x+y+z=1$ on the plane $5x-4y-z=5$ is:

Answer 1

Hint: To get the equation of the projection of the line, we will start from the intersection of line of plane and to get the intersection of line of plane, we will derive it from the given two planes. Then we will find the direction ratio of the intersection line. After that we will get the equation f the projection of the line using the following formula as:
$ \dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}$

Complete step by step solution:
Since, the given planes are:
$\Rightarrow 8x-y-7z=8$ … $\left( i \right)$
And
$\Rightarrow x+y+z=1$ … $\left( ii \right)$
Now, we will try to get the point of intersection of these planes. So, first of all we will consider that $z=0$ . So, the above equation will be as:
$\Rightarrow 8x-y=8$ .. $\left( iii \right)$
And
$\Rightarrow x+y=1$ … $\left( iv \right)$
Now, we will add equation $\left( iii \right)$ and $\left( iv \right)$ as:
$\Rightarrow 8x-y+x+y=8+1$
Now, we will solve the above equation as:
$\Rightarrow 9x=9$
The above equation can be written as:
$\Rightarrow x=\dfrac{9}{9}$
Now, we will have the value of $x$ as:
$\Rightarrow x=1$
Here, we will use this value of $x$ in equation $\left( iv \right)$ as:
$\Rightarrow 1+y=1$ … $\left( iv \right)$
Now, we will place all the numbers one side of equal sign as:
$\Rightarrow y=1-1$
Here, we will get the value of $y$ as:
$\Rightarrow y=0$
Now, we have the intersection point of line as:
$\Rightarrow \left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)=\left( 1,0,0 \right)$
Since, we got the points of intersection. Now, we will have to find the direction ratio of the intersection line. So, let the direction ratio of the intersection of lines is $a,b$ and $c$ .
Since, we already have the two planes, so the direction ratio of the planes is:
The direction ratio of the plane $8x-y-7z=8$ is:
$\Rightarrow \left( 8,-1,-7 \right)$
And the direction ratio of the plane $\Rightarrow x+y+z=1$ is:
$\Rightarrow \left( 1,1,1 \right)$
Since, the direction ratio of plane will be along to the direction ratio of line that is perpendicular to the line. So we will get two equations from the direction ratio as:
$\Rightarrow 8a-b-7c=0$ … $\left( v \right)$
And
$\Rightarrow a+b+c=0$ … $\left( vi \right)$
Now, we will use the equation $\left( v \right)$ and $\left( vi \right)$ to get the value of direction ratio of intersection line as:
$\Rightarrow \dfrac{a}{-1+7}=\dfrac{b}{8+7}=\dfrac{c}{-1-8}$
After solving the above equation we will get:
$\Rightarrow \dfrac{a}{6}=\dfrac{b}{15}=\dfrac{c}{-9}$
Since, we got the required value for getting the equation of the projection of line. Thus, the equation of the line of projection will be as:
$\Rightarrow \dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}$
Now, we will apply the values as:
$\Rightarrow \dfrac{x-1}{6}=\dfrac{y-0}{15}=\dfrac{z-0}{-9}$
After solving the above equation, we will have:
$\Rightarrow \dfrac{x}{6}=\dfrac{y}{15}=\dfrac{z}{-9}$
Here, we will multiply by $3$ in above equation as:
$\Rightarrow 3\times \dfrac{x}{6}=3\times \dfrac{y}{15}=3\times \dfrac{z}{-9}$
Now, we will have the resultant equation as:
$\Rightarrow \dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{-3}$
Hence, the equation of the projection of the line is $\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{-3}$ .

Note: Here, we need to take care of the calculation for direction ratio according to the relation between line and the normal of the plane. We can understand it as:
Here is the diagram that shows the line and the normal of the plane:

Since, the normal of the line is $\hat{n}$ and the direction ratio of the line is $\hat{r}$ in vector form. So,
$\Rightarrow \hat{r}\times \hat{n}=0$
In Cartesian form –
$\Rightarrow {{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}=0$
We will apply this formula for both the given equation and get the value of direction ratio of line.