Question

Find the point of intersection of the lines AB and CD where A(7, -6, 1), B(17, -18, -3), C(1, 4, -5) and D = (3, 4, 11).​

Answer 1

The vector form of equation of the line AB is,

$\longrightarrow\vec{r_1}=\left<7,\ -6,\ 1\right>+\lambda_1\left<7-17,\ (-6)-(-18),\ 1-(-3)\right>$

$\longrightarrow\vec{r_1}=\left<7,\ -6,\ 1\right>+\lambda_1\left<-10,\ 12,\ 4\right>$

$\longrightarrow\vec{r_1}=\left<7-10\lambda_1,\ -6+12\lambda_1,\ 1+4\lambda_1\right>$

and that of line CD is,

$\longrightarrow\vec{r_2}=\left<1,\ 4,\ -5\right>+\lambda_2\left<3-1,\ 4-4,\ 11-(-5)\right>$

$\longrightarrow\vec{r_2}=\left<1,\ 4,\ -5\right>+\lambda_2\left<2,\ 0,\ 16\right>$

$\longrightarrow\vec{r_2}=\left<1+2\lambda_2,\ 4,\ -5+16\lambda_2\right>$

To get the position vector of intersection point, let,

$\longrightarrow\vec{r_1}=\vec{r_2}$

$\longrightarrow\left<7-10\lambda_1,\ -6+12\lambda_1,\ 1+4\lambda_1\right>=\left<1+2\lambda_2,\ 4,\ -5+16\lambda_2\right>$

Equating corresponding 'y' components,

$\longrightarrow-6+12\lambda_1=4$

$\longrightarrow\lambda_1=\dfrac{5}{6}$

Equating corresponding 'x' components and putting value of $\lambda_1,$

$\longrightarrow7-10\lambda_1=1+2\lambda_2$

$\longrightarrow7-10\cdot\dfrac{5}{6}=1+2\lambda_2$

$\longrightarrow7-\dfrac{25}{3}=1+2\lambda_2$

$\longrightarrow\lambda_2=-\dfrac{7}{6}$

But on equating corresponding 'z' components and putting values of $\lambda_1$ and $\lambda_2,$

$\longrightarrow1+4\lambda_1=-5+16\lambda_2$

$\longrightarrow1+4\cdot\dfrac{5}{6}=-5+16\cdot-\dfrac{7}{6}$

$\longrightarrow\dfrac{13}{3}=-\dfrac{71}{3}$

This contradiction implies the lines AB and CD do not intersect each other.

Hence there is no point of intersection.