Question 15 Marks
Show that the lines
$\vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}+4 \hat{k})$ and $\vec{r}=(4 \hat{i}+\hat{j})+\mu(5 \hat{i}+2 \hat{j}+\hat{k})$
intersect. Also, find their point intersection.
$\vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}+4 \hat{k})$ and $\vec{r}=(4 \hat{i}+\hat{j})+\mu(5 \hat{i}+2 \hat{j}+\hat{k})$
intersect. Also, find their point intersection.
Answer
View full question & answer→Here, it is given that
$\vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}+4 \hat{k})$
$\vec{r}=(4 \hat{i}+\hat{j})+\mu(5 \hat{i}+2 \hat{j}+\hat{k})$
Here,
$\overrightarrow{a_1}=i+2 \hat{j}+3 \hat{k}$
$\overrightarrow{b_1}=2 \hat{\imath}+3 \hat{j}+4 \hat{k}$
$\overrightarrow{a_2}=4 \hat{\imath}+\hat{j}$
$\overrightarrow{b_2}=5 \hat{i}+2 \hat{j}+\hat{k}$
Thus,
$\begin{array}{l} \overrightarrow{b_1} \times \overrightarrow{b_2}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 5 & 2 & 1 \end{array}\right| \end{array}$
$=\hat{i}(3-8)-\hat{j}(2-20)+\hat{k}(4-15) $
$\therefore \overrightarrow{b_1} \times \overrightarrow{b_2}=-5 \hat{i}+18 \hat{j}-11 \hat{k}$
$\therefore\left|\overrightarrow{ b _1} \times \overrightarrow{ b _2}\right|=\sqrt{(-5)^2+18^2+(-11)^2}$
$=\sqrt{25+324+121}$
$=\sqrt{470}$
$\overrightarrow{a_2}-\overrightarrow{a_1}=(4-1) \hat{i}+(1-2) \hat{\jmath}+(0-3) \hat{ k }$
$\therefore \overrightarrow{a_2}-\overrightarrow{a_1}=3 \hat{\imath}-\hat{\jmath}-3 \hat{k}$
Now, we have
$\left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right) \cdot\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)=(-5 \hat{i}+18 \hat{j}-11 \hat{k}) \cdot(3 \hat{i}-\hat{j}-3 \hat{k})$
$=((-5) \times 3)+(18 \times(-1))+((-11) \times(-3))$
$=15-18+33$
$= 0$
Thus, the distance between the given lines is
$\begin{array}{l} d =\left|\frac{\left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right) \cdot\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)}{\left|\overrightarrow{b_1} \times \overrightarrow{b_2}\right|}\right| \\ \therefore d =\left|\frac{0} {\sqrt{470}}\right|\end{array}$
$\therefore d = 0$ units
As $d = 0$
Thus, the given lines intersect each other.
Now, to find a point of intersection, let us convert given vector equations into Cartesian equations.
For that putting $\overrightarrow{ r }=x \hat{ i }+y \hat{ j }+z \hat{ k }$ in given equations,
$\Rightarrow \vec{L}_1: x \hat{ i }+ y \hat{ j }+ zk =(i+2 j+3 \hat{k})+\lambda(2 i+3 \hat{j}+4 \hat{k})$
$\Rightarrow \vec{L}_2: x \hat{ i }+ y \hat{ j }+ zk =(4 \hat{i}+\hat{\jmath})+\mu(5 \hat{\imath}+2 \hat{\jmath}+\hat{k})$
$\Rightarrow \vec{L}_1:( x -1) \hat{ i }+( y -2) \hat{ j }+(z-3) \hat{k}=2 \lambda \hat{\imath}+3 \lambda \hat{j}+4 \lambda \hat{k}$
$\Rightarrow \vec{L}_2:( x -4) \hat{ i }+( y -1) \hat{ j }+(z-0) \hat{k}=5 \mu \hat{\imath}+2 \mu \hat{j}+\mu \hat{k}$
$\Rightarrow \vec{L}_1: \frac{ x -1}{2}=\frac{ y -2}{3}=\frac{z-3}{4}=\lambda$
$\therefore \vec{L}_2: \frac{x-4}{5}=\frac{y-1}{2}=\frac{z-0}{1}=\mu$
General point on $L1$ is
$x _1=2 \lambda+1, y _1=3 \lambda+2, z _1=4 \lambda+3$
Suppose, $P\left(x_1, y_1, z_1\right)$be point of intersection of two given lines.
Thus, point $P$ satisfies the equation of line $\vec{L}_2$
$\Rightarrow \frac{2 \lambda+1-4}{5}=\frac{3 \lambda+2-1}{2}=\frac{4 \lambda+3-0}{1}$
$\therefore \frac{2 \lambda-3}{5}=\frac{3 \lambda+1}{2}$
$\Rightarrow 4 \lambda-6=15 \lambda+5$
$\Rightarrow 11 \lambda=-11$
$\Rightarrow \lambda=-1$
Thus, $x_1=2(-1)+1, y_1=3(-1)+2, z_1=4(-1)+3$
$\Rightarrow x_1=-1, y_1=-1, z_1=-1$
Therefore, point of intersection of given lines is $(-1, -1, -1).$
$\vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}+4 \hat{k})$
$\vec{r}=(4 \hat{i}+\hat{j})+\mu(5 \hat{i}+2 \hat{j}+\hat{k})$
Here,
$\overrightarrow{a_1}=i+2 \hat{j}+3 \hat{k}$
$\overrightarrow{b_1}=2 \hat{\imath}+3 \hat{j}+4 \hat{k}$
$\overrightarrow{a_2}=4 \hat{\imath}+\hat{j}$
$\overrightarrow{b_2}=5 \hat{i}+2 \hat{j}+\hat{k}$
Thus,
$\begin{array}{l} \overrightarrow{b_1} \times \overrightarrow{b_2}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 5 & 2 & 1 \end{array}\right| \end{array}$
$=\hat{i}(3-8)-\hat{j}(2-20)+\hat{k}(4-15) $
$\therefore \overrightarrow{b_1} \times \overrightarrow{b_2}=-5 \hat{i}+18 \hat{j}-11 \hat{k}$
$\therefore\left|\overrightarrow{ b _1} \times \overrightarrow{ b _2}\right|=\sqrt{(-5)^2+18^2+(-11)^2}$
$=\sqrt{25+324+121}$
$=\sqrt{470}$
$\overrightarrow{a_2}-\overrightarrow{a_1}=(4-1) \hat{i}+(1-2) \hat{\jmath}+(0-3) \hat{ k }$
$\therefore \overrightarrow{a_2}-\overrightarrow{a_1}=3 \hat{\imath}-\hat{\jmath}-3 \hat{k}$
Now, we have
$\left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right) \cdot\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)=(-5 \hat{i}+18 \hat{j}-11 \hat{k}) \cdot(3 \hat{i}-\hat{j}-3 \hat{k})$
$=((-5) \times 3)+(18 \times(-1))+((-11) \times(-3))$
$=15-18+33$
$= 0$
Thus, the distance between the given lines is
$\begin{array}{l} d =\left|\frac{\left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right) \cdot\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)}{\left|\overrightarrow{b_1} \times \overrightarrow{b_2}\right|}\right| \\ \therefore d =\left|\frac{0} {\sqrt{470}}\right|\end{array}$
$\therefore d = 0$ units
As $d = 0$
Thus, the given lines intersect each other.
Now, to find a point of intersection, let us convert given vector equations into Cartesian equations.
For that putting $\overrightarrow{ r }=x \hat{ i }+y \hat{ j }+z \hat{ k }$ in given equations,
$\Rightarrow \vec{L}_1: x \hat{ i }+ y \hat{ j }+ zk =(i+2 j+3 \hat{k})+\lambda(2 i+3 \hat{j}+4 \hat{k})$
$\Rightarrow \vec{L}_2: x \hat{ i }+ y \hat{ j }+ zk =(4 \hat{i}+\hat{\jmath})+\mu(5 \hat{\imath}+2 \hat{\jmath}+\hat{k})$
$\Rightarrow \vec{L}_1:( x -1) \hat{ i }+( y -2) \hat{ j }+(z-3) \hat{k}=2 \lambda \hat{\imath}+3 \lambda \hat{j}+4 \lambda \hat{k}$
$\Rightarrow \vec{L}_2:( x -4) \hat{ i }+( y -1) \hat{ j }+(z-0) \hat{k}=5 \mu \hat{\imath}+2 \mu \hat{j}+\mu \hat{k}$
$\Rightarrow \vec{L}_1: \frac{ x -1}{2}=\frac{ y -2}{3}=\frac{z-3}{4}=\lambda$
$\therefore \vec{L}_2: \frac{x-4}{5}=\frac{y-1}{2}=\frac{z-0}{1}=\mu$
General point on $L1$ is
$x _1=2 \lambda+1, y _1=3 \lambda+2, z _1=4 \lambda+3$
Suppose, $P\left(x_1, y_1, z_1\right)$be point of intersection of two given lines.
Thus, point $P$ satisfies the equation of line $\vec{L}_2$
$\Rightarrow \frac{2 \lambda+1-4}{5}=\frac{3 \lambda+2-1}{2}=\frac{4 \lambda+3-0}{1}$
$\therefore \frac{2 \lambda-3}{5}=\frac{3 \lambda+1}{2}$
$\Rightarrow 4 \lambda-6=15 \lambda+5$
$\Rightarrow 11 \lambda=-11$
$\Rightarrow \lambda=-1$
Thus, $x_1=2(-1)+1, y_1=3(-1)+2, z_1=4(-1)+3$
$\Rightarrow x_1=-1, y_1=-1, z_1=-1$
Therefore, point of intersection of given lines is $(-1, -1, -1).$
