Question

Find the vector and cartesian equations of line passing through the point (1, 2, -4) and perpendicular to the two lines,
$\dfrac{x-8}{3}=\dfrac{y+19}{-16}=\dfrac{z-10}{7}$ and $\dfrac{x-15}{3}=\dfrac{y-29}{8}=\dfrac{z-5}{-5}$.

Answer 1

Hint: From the equation of the Cartesian plane using the point (1, 2, -4), form the equation of vectors from the 3 lines formed. Thus $\overrightarrow{{{b}_{1}}}\bot \overrightarrow{{{b}_{2}}}$ and $\overrightarrow{{{b}_{1}}}\bot \overrightarrow{{{b}_{3}}}$. Find the equations, get the value and substitute in the vector equation.

Complete step-by-step answer:
We have to find the vector and Cartesian equation of the line passing through the point (1, 2, -4).
Let the Cartesian equation of a line passing through a point be given as,
$\dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}$.
$\therefore \left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)=\left( 1,2,-4 \right)$, thus substitute the value in the above equation.
So the Cartesian equation of the line passing through (1, 2, -4) is
$\dfrac{x-1}{a}=\dfrac{y-2}{b}=\dfrac{z-4}{c}......(1)$
The other line given as,

& \dfrac{x-8}{3}=\dfrac{y+19}{-16}=\dfrac{z-10}{7}......(2) \\\ & \dfrac{x-15}{3}=\dfrac{y-29}{8}=\dfrac{z-5}{-5}......(3) \\\ \end{aligned}$$ The parallel vectors $$\overrightarrow{{{b}_{1}}},\overrightarrow{{{b}_{2}}},\overrightarrow{{{b}_{3}}}$$ of equation (1), (2) and (3) can be given as, $$\overrightarrow{{{b}_{1}}}=a\hat{i}+b\hat{j}+c\hat{k}$$ from equation (1) $$\overrightarrow{{{b}_{2}}}=3\hat{i}-16\hat{j}+7\hat{k}$$ from equation (2) $$\overrightarrow{{{b}_{3}}}=3\hat{i}+8\hat{j}-5\hat{k}$$ from equation (3) It is said that the Cartesian equation of line passing through the point are perpendicular to the two lines. Thus we can say that $$\overrightarrow{{{b}_{1}}}$$ is perpendicular to $$\overrightarrow{{{b}_{2}}}$$ and $$\overrightarrow{{{b}_{1}}}$$ is perpendicular to $$\overrightarrow{{{b}_{3}}}$$, i.e. $$\overrightarrow{{{b}_{1}}}\bot \overrightarrow{{{b}_{2}}}$$ and $$\overrightarrow{{{b}_{1}}}\bot \overrightarrow{{{b}_{3}}}$$. As $$\overrightarrow{{{b}_{1}}}\bot \overrightarrow{{{b}_{2}}}$$ we can say that $$\overrightarrow{{{b}_{1}}}\bot \overrightarrow{{{b}_{2}}}=0$$. Similarly, $$\overrightarrow{{{b}_{1}}}\bot \overrightarrow{{{b}_{3}}}$$ we can say that $$\overrightarrow{{{b}_{1}}}\bot \overrightarrow{{{b}_{3}}}=0$$. Thus let us take the dot product of these vectors. $$\begin{aligned} & \overrightarrow{{{b}_{1}}}.\overrightarrow{{{b}_{2}}}=0 \\\ & \left( a\hat{i}+b\hat{j}+c\hat{k} \right).\left( 3\hat{i}-16\hat{j}+7\hat{k} \right)=0 \\\ & \Rightarrow 3a-16b+7c=0......(4) \\\ \end{aligned}$$ Similarly, $$\begin{aligned} & \overrightarrow{{{b}_{1}}}.\overrightarrow{{{b}_{3}}}=0 \\\ & \left( a\hat{i}+b\hat{j}+c\hat{k} \right).\left( 3\hat{i}+8\hat{j}-5\hat{k} \right)=0 \\\ & \Rightarrow 3a+8b-5c=0.......(5) \\\ \end{aligned}$$ Thus we have equation (4)and (5). $$\begin{aligned} & 3a-16b+7c=0 \\\ & 3a+8b-5c=0 \\\ \end{aligned}$$ From this we can say that, $$\dfrac{a}{\left( -16\times -5 \right)-\left( 8\times 7 \right)}=\dfrac{b}{\left( 3\times 7 \right)-\left( 3\times -5 \right)}=\dfrac{c}{\left( 3\times 8 \right)-\left( 3\times -16 \right)}$$ $$\Rightarrow \dfrac{a}{80-56}=\dfrac{b}{21+15}=\dfrac{c}{24+48}$$ $$\Rightarrow \dfrac{a}{24}=\dfrac{b}{36}=\dfrac{c}{72}$$ Multiply throughout by 12 and simplify it. $$\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{6}=\lambda $$ Thus we can say that $$\dfrac{a}{2}=\lambda $$, $$\dfrac{b}{3}=\lambda $$ and $$\dfrac{c}{6}=\lambda $$. $$\therefore a=2\lambda ,b=3\lambda ,c=6\lambda $$. Now let us put these values in equation (1). $$\dfrac{x-1}{2\lambda }=\dfrac{y-2}{3\lambda }=\dfrac{z+4}{6\lambda }$$ Cancel the $$\lambda $$ in the denominator, we get, $$\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z+4}{6}$$. Hence we get the vector equation, i.e. $$\overrightarrow{r}=a+\lambda u$$. $$\overrightarrow{r}=\left( \hat{i}+2\hat{j}-4\hat{k} \right)+\lambda \left( 2\hat{i}+3\hat{j}+6\hat{k} \right)$$. Note: If two vectors are perpendicular then their dot product is equal to zero. This is one of the most important points in the vector, which we have used here to calculate the required equation of the vector by taking a dot product. Thus remember these points while solving vector equations.