Question

The point of intersection of the line joining the points $(3,4,1)$ and $(5,1,6)$ and the $xy$ plane is

• A. $(13,23,0)$
• B. $\left(\frac{13}{5},\frac{23}{5},0\right)$
• C. $(-13,23,0)$
• D. $\left(-\frac{13}{5},\frac{23}{5},0\right)$

Answer 1

The correct option is B

$\left(\frac{13}{5},\frac{23}{5},0\right)$

Explanation for the correct answer:

Line and plane intersection:

Equation of line passing via $\left(a_1,a_2,a_3\right)and(b_1,b_2,b_3)$

$\frac{\mathbf{x}\mathbf{-}{\mathbf{a}}_{\mathbf{1}}}{{\mathbf{b}}_{\mathbf{1}}\mathbf{-}{\mathbf{a}}_{\mathbf{1}}}\mathbf{=}\frac{\mathbf{x}\mathbf{-}{\mathbf{a}}_{\mathbf{2}}}{{\mathbf{b}}_{\mathbf{2}}\mathbf{-}{\mathbf{a}}_{\mathbf{2}}}\mathbf{=}\frac{\mathbf{x}\mathbf{-}{\mathbf{a}}_{\mathbf{3}}}{{\mathbf{b}}_{\mathbf{3}}\mathbf{-}{\mathbf{a}}_{\mathbf{3}}}$

The equation of the line joining the points $(3,4,1)$ and $(5,1,6)$ can be written as

$\frac{x-3}{5-3}=\frac{y-4}{1-4}=\frac{z-1}{6-1}$

$\Rightarrow$$\frac{x-3}{2}=\frac{y-4}{-3}=\frac{z-1}{5}=\lambda$ (say)

$\Rightarrow$$x=2\lambda +3,y=-3\lambda +4,z=5\lambda +1$

For $xy-$plane,

$z=0$

$$\begin{gathered} \Rightarrow 5\lambda +1=0 \\ \Rightarrow \lambda =-\frac{1}{5} \end{gathered}$$

Substituting this value of $\lambda$ in co-ordinates of the point we get

$$\begin{gathered} \therefore x=2\left(\frac{-1}{5}\right)+3=\frac{15-2}{5}=\frac{13}{5} \\ \therefore y=-3\left(\frac{-1}{5}\right)+4=\frac{3+20}{5}=\frac{23}{5} \end{gathered}$$

Hence, the point of intersection of the line joining the points $(3,4,1)$ and $(5,1,6)$ and the $xy-$ plane is $\left(\frac{13}{5},\frac{23}{5},0\right)$

Hence, option B is the correct answer.