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# Find the distance of point $\left( 1,-5,9 \right)$ from plane $x-y+z=5$ measured along the line $x=y=z$.

Last updated date: 20th Jun 2024
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Hint: In this question, we first need to find the equation of the line parallel to given line and passing through the given point which is given by the formula $\dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}=r$. Then we get the general point on that line parallel to $x=y=z$. Now, we need to find the point of intersection of this point and the plane which gives the value of r and so the point. Now, we need to find the distance between this point and the given point in the question using the formula $\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{z}_{2}}-{{z}_{1}} \right)}^{2}}}$

Equation of a line passing through a fixed point $A\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$and having direction ratios a, b, c is given by
$\dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}=r$
Now, we need to find the line parallel to $x=y=z$ and passing through $\left( 1,-5,9 \right)$
Here, the direction ratios are given by
$a=1,b=1,c=1$
Now, on comparing the given point we get,
${{x}_{1}}=1,{{y}_{1}}=-5,{{z}_{1}}=9$
Let us now substitute the respective in the above equation and simplify further
$\Rightarrow \dfrac{x-1}{1}=\dfrac{y-\left( -5 \right)}{1}=\dfrac{z-9}{1}=r$
Now, this can be further written as
$\Rightarrow \dfrac{x-1}{1}=\dfrac{y+5}{1}=\dfrac{z-9}{1}=r$
Now, the general point on this line can be given by
$\Rightarrow x-1=r,y+5=r,z-9=r$
Now, on rearranging the terms this can be further written as
$\Rightarrow \left( x,y,z \right)=\left( r+1,r-5,r+9 \right)$
Now, let us find the point of intersection of this line and the given plane
Here, we get the point of intersection by substituting the general form of point in the plane equation.
$\Rightarrow x-y+z=5$

Now, on substituting the general point in this equation we get,
$\Rightarrow r+1-\left( r-5 \right)+r+9=5$
Now, this can be further written as
$\Rightarrow 2r+10-r+5=5$
Now, on further simplification we get,
$\therefore r=-10$
Let us now substitute this value in the general point to get the point of intersection
$\Rightarrow \left( x,y,z \right)=\left( r+1,r-5,r+9 \right)$
Now, on substituting the respective value we get,
$\Rightarrow \left( x,y,z \right)=\left( -10+1,-10-5,-10+9 \right)$
Now, on simplifying this further we get,
$\therefore \left( x,y,z \right)=\left( -9,-15,-1 \right)$
Let us assume this point of intersection as A
$\therefore A\left( -9,-15,-1 \right)$
Now, the distance between the point $\left( 1,-5,9 \right)$from the plane $x-y+z=5$along the given line will be the distance between the points $A\left( -9,-15,-1 \right)$ and $\left( 1,-5,9 \right)$
As we already know that the distance between two points is given by the formula
$\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{z}_{2}}-{{z}_{1}} \right)}^{2}}}$
Now, on substituting the respective values we get,
$\Rightarrow \sqrt{{{\left( 1-\left( -9 \right) \right)}^{2}}+{{\left( -5-\left( -15 \right) \right)}^{2}}+{{\left( 9-\left( -1 \right) \right)}^{2}}}$
Now, this can be further written in the simplified form as
$\Rightarrow \sqrt{{{10}^{2}}+{{10}^{2}}+{{10}^{2}}}$
Now, on further simplification we get,
$\Rightarrow \sqrt{3\left( {{10}^{2}} \right)}$
Now, this can be further written as
$\Rightarrow 10\sqrt{3}$

Note:
It is important to note that to find the distance along the line $x=y=z$we need to find a line that is parallel to this line and passing through the given point. Now, finding the point of intersection of this line with the plane helps us in finding the required distance.
It is also to be noted that while finding the equation of a line or any point we need to substitute the respective values without interchanging. Also, while calculating we should not neglect any of the terms and consider the incorrect sign because it changes the final result.