Question

Find the equation of the line passing through the point $\left( 1,2,-4 \right)$ and perpendicular to two lines $\overrightarrow{r}=8\widehat{i}-19\widehat{j}+10\widehat{k}+\lambda \left( 3\widehat{i}-16\widehat{j}+7\widehat{k} \right)\text{ and }\overrightarrow{r}=15\widehat{i}+29\widehat{j}+5\widehat{k}+\mu \left( 3\widehat{i}+8\widehat{j}-5\widehat{k} \right)$.

Answer 1

To solve this question, we should convert the given lines into Cartesian form and obtain the directional ratios of the two lines. After doing this step, we should use the perpendicular property in the lines as given in the question. A line written in the form of $\text{ }\overrightarrow{r}={{x}_{1}}\widehat{i}+{{y}_{1}}\widehat{j}+{{z}_{1}}\widehat{k}+\mu \left( a\widehat{i}+b\widehat{j}+c\widehat{k} \right)$ can be written in Cartesian form as $\dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}$ where a, b, c are the directional ratios of the line. Two lines with the directional ratios $a,b,c\text{ and }d,e,f$ are perpendicular when $ad+be+cf=0$. By assuming the required line’s directional ratios as a, b, c and using the above perpendicular condition, we get the directional ratios of the line. The equation of the line passing through the point $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and having directional ratios as a, b, c is given as $\dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}$. Using this, we can get the equation of the line. We can also convert the final equation of the line into the vector format using the above relation.

Complete step by step answer:
A line written in the form of $\text{ }\overrightarrow{r}={{x}_{1}}\widehat{i}+{{y}_{1}}\widehat{j}+{{z}_{1}}\widehat{k}+\mu \left( a\widehat{i}+b\widehat{j}+c\widehat{k} \right)$ can be written in Cartesian form as $\dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}$ where a, b, c are the directional ratios of the line.
Using this, we can get the Cartesian equations of the given lines.
$\overrightarrow{r}=8\widehat{i}-19\widehat{j}+10\widehat{k}+\lambda \left( 3\widehat{i}-16\widehat{j}+7\widehat{k} \right)$ can be written as
$\dfrac{x-8}{3}=\dfrac{y+19}{-16}=\dfrac{z-10}{7}$
The directional ratios of the line are $3,-16,7$
$\overrightarrow{r}=15\widehat{i}+29\widehat{j}+5\widehat{k}+\mu \left( 3\widehat{i}+8\widehat{j}-5\widehat{k} \right)$ can be written as
$\dfrac{x-15}{3}=\dfrac{y-29}{8}=\dfrac{z-5}{-5}$
The directional ratios of the line are $3,8,-5$
Let the directional ratios of the required line be $a,b,c$ respectively. It is given in the question that the required line is perpendicular to both the given lines
Two lines with the directional ratios $a,b,c\text{ and }d,e,f$ are perpendicular when $ad+be+cf=0$.
Using this property with the required line and the first given line, we get
$3a-16b+7c=0\to \left( 1 \right)$
Using this property with the required line and the second given line, we get
$3a+8b-5c=0\to \left( 2 \right)$
Here, we have three unknowns a, b, c but only two equations. To make the equations into two variables, we can divide the whole equation by c.
Dividing by c, we get
$\begin{aligned} & 3\dfrac{a}{c}-16\dfrac{b}{c}=-7 \\\ & 3\dfrac{a}{c}+8\dfrac{b}{c}=5 \\\ \end{aligned}$
Subtracting the two equations, we get
$\begin{aligned} & -24\dfrac{b}{c}=-12 \\\ & \dfrac{b}{c}=\dfrac{1}{2} \\\ & b=\dfrac{c}{2} \\\ \end{aligned}$
Substituting $\dfrac{b}{c}=\dfrac{1}{2}$ in any of the above equations in a, b, we get

& 3\dfrac{a}{c}+8\times \dfrac{1}{2}=5 \\\ & 3\dfrac{a}{c}=1 \\\ & \dfrac{a}{c}=\dfrac{1}{3} \\\ \end{aligned}$$ $a=\dfrac{c}{3}$ So, the directional ratios of the required line are $\dfrac{c}{3},\dfrac{c}{2},c$ We know a property of ratios that $a:b:c=ka:kb:kc$ We can multiply or divide by any number to all the directional ratios and their value doesn’t change. Multiplying all the directional ratios by $\dfrac{6}{c}$, we get $\begin{aligned} & \dfrac{c}{3}\times \dfrac{6}{c},\dfrac{c}{2}\times \dfrac{6}{c},c\times \dfrac{6}{c} \\\ & \text{Directional ratios}=2,3,6 \\\ \end{aligned}$ It is given that the line passes through the point $\left( 1,2,-4 \right)$. The equation of the line passing through the point $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and having directional ratios as a, b, c is given as $\dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}$. Using this property with $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)=\left( 1,2,-4 \right)$ and $\left( a,b,c \right)=\left( 2,3,6 \right)$, we get $\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z+4}{6}$ **$\therefore $ The required line is $\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z+4}{6}$** **Note:** Some students think that there are three variables a, b, c which are directional ratios and we are given only two equations and there is insufficient information. The name itself suggests that the three variables are ratios. This means that their relative value is needed but not the actual value. That is why we can take one of the values as 1 and get the other values and at the end we can multiply them with any number to remove the fractions.