Question

Find the coordinates of the point where the line through the points A = (3,4,1) and B = (5,1,6) crosses the $xy$- plane.

Answer 1

We start solving this problem by considering the vector form of the given two points. Then we use the formula for the equation of line passing through two points with position vectors $\overrightarrow{a}$and $\overrightarrow{b}$ i.e., $\overrightarrow{r}=\overrightarrow{a}+\lambda \left( \overrightarrow{b}-\overleftrightarrow{a} \right)$. Then we consider the general line which crosses the $xy$- plane and equate it to the obtained line equation. Then we equate the corresponding components to get the value of $\lambda$. Hence, we get the coordinates of the required point.

Complete step by step answer:
Let us consider the given points A = (3,4,1) and B = (5,1,6).
Now, we change them into vector form, we get
$\overrightarrow{a}=3\hat{i}+4\hat{j}+\hat{k}$ and $\overrightarrow{b}=5\hat{i}+\hat{j}+6\hat{k}$.
Let us consider the formula for the equation of line passing through two points with position vectors $\overrightarrow{a}$and $\overrightarrow{b}$, $\overrightarrow{r}=\overrightarrow{a}+\lambda \left( \overrightarrow{b}-\overleftrightarrow{a} \right)$.
Using the above formula, we get

& \overrightarrow{r}=\left( 3\hat{i}+4\hat{j}+\hat{k} \right)+\lambda \left[ \left( 5\hat{i}+\hat{j}+6\hat{k} \right)-\left( 3\hat{i}+4\hat{j}+\hat{k} \right) \right] \\\ & \overrightarrow{r}=\left( 3\hat{i}+4\hat{j}+\hat{k} \right)+\lambda \left[ 5\hat{i}+\hat{j}+6\hat{k}-3\hat{i}-4\hat{j}-\hat{k} \right] \\\ & \overrightarrow{r}=\left( 3\hat{i}+4\hat{j}+\hat{k} \right)+\lambda \left[ \left( 5-3 \right)\hat{i}+\left( 1-4 \right)\hat{j}+\left( 6-1 \right)\hat{k} \right] \\\ & \overrightarrow{r}=\left( 3\hat{i}+4\hat{j}+\hat{k} \right)+\lambda \left[ 2\hat{i}-3\hat{j}+5\hat{k} \right] \\\ & \overrightarrow{r}=\left( 3+2\lambda \right)\hat{i}+\left( 4-3\lambda \right)\hat{j}+\left( 1+5\lambda \right)\hat{k} \\\ \end{aligned}$$ So, we get the general equation of line passing through two points A = (3,4,1) and B = (5,1,6) as $$\overrightarrow{r}=\left( 3+2\lambda \right)\hat{i}+\left( 4-3\lambda \right)\hat{j}+\left( 1+5\lambda \right)\hat{k}.....................\left( 1 \right)$$ We were given that the above line crosses $xy$- plane. So, we consider the $z$- coordinate of the required point as 0. Let us consider the coordinates of the point where the line crosses the $xy$- plane as $\left( x,y,0 \right)$. So, we get $\overrightarrow{r}=x\hat{i}+y\hat{j}+0\hat{k}....................\left( 2 \right)$ Since this point crosses the given plane, we substitute equation (2) in equation (1), we get $x\hat{i}+y\hat{j}+0\hat{k}=\left( 3+2\lambda \right)\hat{i}+\left( 4-3\lambda \right)\hat{j}+\left( 1+5\lambda \right)\hat{k}$ Now, let us equate the corresponding components. Then, we get $\begin{aligned} & x=\left( 3+2\lambda \right)................\left( 3 \right) \\\ & y=\left( 4-3\lambda \right)...............\left( 4 \right) \\\ & 0=\left( 1+5\lambda \right).................\left( 5 \right) \\\ \end{aligned}$ Now, we solve equation (5), we get $\begin{aligned} & -5\lambda =1 \\\ & \lambda =-\dfrac{1}{5} \\\ \end{aligned}$ Now, by substituting the value of $\lambda $ in equation (3), we get $\begin{aligned} & x=3+2\lambda \\\ & \Rightarrow x=3+2\left( -\dfrac{1}{5} \right) \\\ & \Rightarrow x=3-\dfrac{2}{5} \\\ & \Rightarrow x=\dfrac{13}{5} \\\ \end{aligned}$ And by substituting the value of $\lambda $ in equation (4), we get $\begin{aligned} & y=4-3\lambda \\\ & \Rightarrow y=4-3\left( -\dfrac{1}{5} \right) \\\ & \Rightarrow y=4+\dfrac{3}{5} \\\ & \Rightarrow y=\dfrac{23}{5} \\\ \end{aligned}$ Therefore, the coordinates of the point where line through the points A = (3,4,1) and B = (5,1,6) crosses the $xy$- plane are $\left( \dfrac{13}{5},\dfrac{23}{5},0 \right)$. Hence, the final answer is $\left( \dfrac{13}{5},\dfrac{23}{5},0 \right)$. **Note:** There is a possibility of making a mistake by considering $x$and $y$ coordinates as zero as the line crosses $xy$ -plane. But it is wrong, when a line crosses the xy plane we must consider $z$ - coordinate as zero.