Question

Find the distance of the point P (– 1, – 5, – 10) from the point of intersection of the line joining the point A (2, – 1, 2) and B (5, 3, 4) with the plane x – y + z = 5.

Answer 1

First of all find the equation of the line joining two points $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ by using $\dfrac{\left( x-{{x}_{1}} \right)}{\left( {{x}_{2}}-{{x}_{1}} \right)}=\dfrac{\left( y-{{y}_{1}} \right)}{\left( {{y}_{2}}-{{y}_{1}} \right)}=\dfrac{\left( z-{{z}_{1}} \right)}{\left( {{z}_{2}}-{{z}_{1}} \right)}=k$. Now put the general point of the line in the plane to get the exact point of intersection. Now, find the distance between this point and the given point using $\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{z}_{2}}-{{z}_{1}} \right)}^{2}}}$.

Complete Step-by-step answer:
In this question, we have to find the distance of the point P (– 1, – 5, – 10) from the point of intersection of the line joining the point A (2, – 1, 2) and B (5, 3, 4) with the plane x – y + z = 5. First of all, let us find the equation of the line joining the point A (2, – 1, 2) and B (5, 3, 4).

We know that the equation of any line joining the points $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ is given by:
$\dfrac{\left( x-{{x}_{1}} \right)}{\left( {{x}_{2}}-{{x}_{1}} \right)}=\dfrac{\left( y-{{y}_{1}} \right)}{\left( {{y}_{2}}-{{y}_{1}} \right)}=\dfrac{\left( z-{{z}_{1}} \right)}{\left( {{z}_{2}}-{{z}_{1}} \right)}=k$
So, by considering $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)=\left( 2,-1,2 \right)$ and $\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)=\left( 5,3,4 \right)$, we get the equation of the line as:
$\dfrac{x-2}{5-2}=\dfrac{y-\left( -1 \right)}{3-\left( -1 \right)}=\dfrac{z-2}{4-2}=k$
$\dfrac{x-2}{3}=\dfrac{y+1}{4}=\dfrac{z-2}{2}=k$
So, we get any general point on the line as

& x=3k+2 \\\ & y=4k-1 \\\ & z=2k+2 \\\ \end{aligned}$$ Now, we have to find the point of intersection of this line to the plane x – y + z = 5. So, we will substitute this general point on the line in the plane.  So, by substituting $$\left( x,y,z \right)=\left( 3k+2,4k-1,2k+2 \right)$$ in the plane x – y + z = 5. We get, $$\left( 3k+2 \right)-\left( 4k-1 \right)+2k+2=5$$ $$3k+2-4k+1+2k+2=5$$ $$\left( 5k-4k \right)+5=5$$ k = 0 So by substituting k = 0, we get, (x, y, z) = (2, – 1, 2). Now, we have to find the distance of this point from point P (– 1, – 5, – 10).  We know that the distance between the points $$\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$$ and $$\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$$ is given by: $$\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{z}_{2}}-{{z}_{1}} \right)}^{2}}}$$ So, we get the distance between the points (– 1, – 5, – 10) and (2, – 1, 2) as $$d=\sqrt{{{\left( 2-\left( -1 \right) \right)}^{2}}+{{\left( -1-\left( -5 \right) \right)}^{2}}+{{\left( 2-\left( -10 \right) \right)}^{2}}}$$ So, we get, $$d=\sqrt{{{\left( 3 \right)}^{2}}+{{\left( 4 \right)}^{2}}+{{\left( 12 \right)}^{2}}}$$ = 13 units **So, we get the distance between point P (– 1, – 5, – 10) and point of intersection of the line joining the points A (2, – 1, 2) and B (5, 3, 4) with the plane x – y + z = 5 as 13 units.** **Note:** In this question, as we can see that the point of intersection of the line and plane is point A only which is already given. So, in these types of questions, it is advisable to first check the given points by substituting them in the plane. If they would satisfy the equation of the plane, the steps in finding the equation of the line would be saved. Though the most general method to these types of questions is as done above.