Question

Consider a line $\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z-2}{4}$ and a plane $x+2y+3z=15$ then find the distance of origin from the point of intersection of line and plane
(1) $\dfrac{5}{2}$
(2) 4
(3) $\dfrac{1}{2}$
(4) $\dfrac{9}{2}$

Answer 1

When the line intersects the plane, the general coordinates of a point on the plane must satisfy the equation of the line. First of all, assume $\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z-2}{4}=t$ . Now, calculate the x, y, and z in terms of t. Substitute x, y, and z in terms of t in the equation of the line $x+2y+3z=15$ and get the value of t. Now, using the value of t, get the coordinate of the point of intersection of the line and the plane. At last, use the distance formula, Distance = $\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}+{{\left( {{z}_{1}}-{{z}_{2}} \right)}^{2}}}$
and calculate the distance of origin from the point of intersection of line and plane.

Complete step by step answer:
According to the question, we are given that
The equation of the plane = $\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z-2}{4}$ ……………………………………….(1)
The equation of the line = $x+2y+3z=15$ ……………………………………….(2)
Let us assume that $\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z-2}{4}=t$ ………………………………………..(3)
For the general coordinate of a point lying on the plane,

& \Rightarrow \dfrac{x-1}{2}=t \\\ & \Rightarrow x-1=2t \\\ \end{aligned}$$ $$\Rightarrow x=1+2t$$ The x-coordinate of the point lying on the plane = $$2t+1$$ …………………………………………(4) $$\begin{aligned} & \Rightarrow \dfrac{y+1}{3}=t \\\ & \Rightarrow y+1=3t \\\ \end{aligned}$$ $$\Rightarrow y=3t-1$$ The y-coordinate of the point lying on the plane = $$3t-1$$ ……………………………………………(5) $$\begin{aligned} & \Rightarrow \dfrac{z-2}{4}=t \\\ & \Rightarrow z-2=4t \\\ \end{aligned}$$ $$\Rightarrow z=4t+2$$ The z-coordinate of the point lying on the plane = $$4t+2$$ ……………………………………………(6) From equation (4), equation (5), and equation (6), we have The general coordinate of a point lying on the given plane is $$\left( 2t+1,3t-1,4t+2 \right)$$ …………………………………(7) We are asked to find the point of intersection of the line and the plane. When the line intersects the plane, the general coordinates of a point on the plane must satisfy the equation of the line …………………………………….(8) Now, from equation (7) and equation (8), and on replacing x by $$2t+1$$ , y by $$3t-1$$ , and z by $$4t+2$$ in equation (2), we get $$\begin{aligned} & \Rightarrow \left( 2t+1 \right)+2\left( 3t-1 \right)+3\left( 4t+2 \right)=15 \\\ & \Rightarrow 2t+6t+12t+1-2+6=15 \\\ & \Rightarrow 20t+5=15 \\\ & \Rightarrow 20t=15-5 \\\ & \Rightarrow 20t=10 \\\ & \Rightarrow t=\dfrac{10}{20} \\\ \end{aligned}$$ $$\Rightarrow t=\dfrac{1}{2}$$ …………………………………………(9) Now, replacing the value of t by $$\dfrac{1}{2}$$ in equation (7), we get The coordinate of the point of intersection of line and the given plane = $$\left( 2\times \dfrac{1}{2}+1,3\times \dfrac{1}{2}-1,4\times \dfrac{1}{2}+2 \right)=\left( 2,\dfrac{1}{2},4 \right)$$ …………………………………………….(10) We are asked to find the distance of the point of intersection of line and the given plane. The coordinate of origin = $$\left( 0,0,0 \right)$$ ………………………………….(11) The coordinate of the point of intersection of line and the given plane = $$\left( 2,\dfrac{1}{2},4 \right)$$ ……………………………….(12) Using the distance formula, Distance = $$\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}+{{\left( {{z}_{1}}-{{z}_{2}} \right)}^{2}}}$$ …………………………………(13) From equation (11), equation (12), and equation (13), we get Distance between the origin and point of intersection of line and plane = $$\sqrt{{{\left( 0-2 \right)}^{2}}+{{\left( 0-\dfrac{1}{2} \right)}^{2}}+{{\left( 0-4 \right)}^{2}}}=\sqrt{4+\dfrac{1}{4}+16}=\sqrt{\dfrac{81}{4}}=\dfrac{9}{2}$$ . Therefore, the distance of origin from the point of intersection of line and plane is $$\dfrac{9}{2}$$ . **So, the correct answer is “Option D”.** **Note:** In this question, one might face difficulty to find out the coordinate of the point of intersection of line and plane. The best way to get the coordinate of the point of intersection of line and plane is to assume the equation of the line as $$\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z-2}{4}=t$$ and then calculate the x-coordinate, y-coordinate and z-coordinate in terms of $$t$$ . Now, with the help of t get the coordinate of the point of intersection of line and plane.