Question

If a point $R(4,y,z)$ lies on the line segment joining the points $P(2,-3,4)$ and $Q\left( 8,0,10 \right)$, then the distance of R from the origin is
\[\begin{align}
  & a)2\sqrt{14} \\
 & b)6 \\
 & c)\sqrt{53} \\
 & d)2\sqrt{21} \\
 & \\
\end{align}\]

Answer 1

Hint:Now we know that the equation of line passing through $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ is given by $\dfrac{x-{{x}_{1}}}{{{x}_{1}}-{{x}_{2}}}=\dfrac{y-{{y}_{1}}}{{{y}_{1}}-{{y}_{2}}}=\dfrac{z-{{z}_{1}}}{z-{{z}_{2}}}$ . Hence using this we can find the equation of line passing through points $P(2,-3,4)$ and $Q\left( 8,0,10 \right)$ . Now since we know the equation of line we can find any general point on line. Hence we can compare the general point with R and find the coordinates of R. Now we can easily calculate the distance of R from origin with the help of distance formula.

Complete step by step answer:
Now consider the points $P(2,-3,4)$ and $Q\left( 8,0,10 \right)$ .
Now, we want to find the equation of line passing though the above points.
Now we have the equation of line passing through $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ is given by $\dfrac{x-{{x}_{1}}}{{{x}_{1}}-{{x}_{2}}}=\dfrac{y-{{y}_{1}}}{{{y}_{1}}-{{y}_{2}}}=\dfrac{z-{{z}_{1}}}{z-{{z}_{2}}}$.
Hence the equation of line passing through $P(2,-3,4)$ and $Q\left( 8,0,10 \right)$ is
$\begin{align}
  & \dfrac{x-2}{2-\left( 8 \right)}=\dfrac{y-(-3)}{-3-0}=\dfrac{z-4}{4-10} \\
 & \Rightarrow \dfrac{x-2}{-6}=\dfrac{y+3}{-3}=\dfrac{z-4}{-6} \\
\end{align}$
Now let us say $\dfrac{x-2}{-6}=\dfrac{y+3}{-3}=\dfrac{z-4}{-6}=\lambda $
Hence we have $x=-6\lambda +2,y=-3\lambda -3,z=-6\lambda +4$
Now we can say that any point on this line will look like $\left( -6\lambda +2,-3\lambda -3,-6\lambda +4 \right)$
Now we know that point R also lies on this line.
Hence R will have similar representation.
Let us say for some ${{\lambda }_{1}}$ we have $R\left( 4,y,z \right)=\left( -6{{\lambda }_{1}}+2,-3{{\lambda }_{1}}-3,-6{{\lambda }_{1}}+4 \right)................\left( 1 \right)$
Now comparing we get
\[\begin{align}
  & -6{{\lambda }_{1}}+2=4 \\
 & \Rightarrow -6{{\lambda }_{1}}=2 \\
\end{align}\]
Dividing the equation by – 6 we get \[{{\lambda }_{1}}=\dfrac{2}{-6}=-\dfrac{1}{3}\]
Now substituting the value of \[{{\lambda }_{1}}\] in equation (1) we get the point R is
$\begin{align}
  & \left( -6\left( -\dfrac{1}{3} \right)+2,-3\left( -\dfrac{1}{3} \right)-3,-6\left( -\dfrac{1}{3} \right)+4 \right) \\
 & \Rightarrow \left( 2+2,1-3,2+4 \right) \\
 & \Rightarrow \left( 4,-2,6 \right) \\
\end{align}$
Hence we get the point R is (4, - 2, 6).
Now we know coordinate of origin is (0, 0, 0).
Now distance between two points $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ is given by $\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{z}_{2}}-{{z}_{1}} \right)}^{2}}}$
Hence the distance between R and origin is
\[\begin{align}
  & \sqrt{{{\left( 4-0 \right)}^{2}}+{{\left( -2-0 \right)}^{2}}+{{\left( 6-0 \right)}^{2}}} \\
 & =\sqrt{16+4+36} \\
 & =\sqrt{56} \\
 & =2\sqrt{14} \\
\end{align}\]
Hence the distance of point R (4, -2, 6) from origin is $2\sqrt{14}$
Option a is the correct option.

Note:
 Now we can also solve this question by another method. Now we know that R lies on the line joining P and Q. Hence with two-point formula, we can write 3 equations of lines which are PQ, PR, and RQ. Now, these 3 equations must be the same as they represent the same line hence we can find y and z. Once we have R we will use the distance formula for finding distance between R and origin.