Question

The line of intersection of the planes $\overset{\to }{\mathop{r}}\,\cdot \left( 3\widehat{i}-\widehat{j}+\widehat{k} \right)=1\text{ and }\overset{\to }{\mathop{r}}\,\cdot \left( \widehat{i}+4\widehat{j}-2\widehat{k} \right)=2$ is$\begin{aligned} & \text{A) }\dfrac{x-\dfrac{6}{13}}{2}=\dfrac{y-\dfrac{5}{13}}{-7}=\dfrac{z}{-13} \\\ & \text{B) }\dfrac{x-\dfrac{6}{13}}{2}=\dfrac{y-\dfrac{5}{13}}{7}=\dfrac{z}{-13} \\\ & \text{C) }\dfrac{x-\dfrac{4}{7}}{-2}=\dfrac{y}{7}=\dfrac{z-\dfrac{5}{7}}{-13} \\\ & \text{D) }\dfrac{x-\dfrac{4}{7}}{2}=\dfrac{y}{-7}=\dfrac{z+\dfrac{5}{7}}{13} \\\ \end{aligned}$

Answer 1

In this to find the common intersecting line of two planes. First we convert the vector form of planes into Cartesian form. After converting into Cartesian form we will substitute one of the coordinates of the three dimensional system as some constant to convert it into two dimension system. By obtaining an equation of planes in a two dimensional system we can this equation simultaneously and obtain the parametric form of equation of lines.

Complete step by step answer:
The equation planes in vector form are $\overset{\to }{\mathop{r}}\,\cdot \left( 3\widehat{i}-\widehat{j}+\widehat{k} \right)=1\text{ and }\overset{\to }{\mathop{r}}\,\cdot \left( \widehat{i}+4\widehat{j}-2\widehat{k} \right)=2$
The direction cosines of the plane $\overset{\to }{\mathop{r}}\,\cdot \left( 3\widehat{i}-\widehat{j}+\widehat{k} \right)=1$ are (3, -1, 1) and perpendicular distance from origin is 1. Therefore the Cartesian equation of planes $\overset{\to }{\mathop{r}}\,\cdot \left( 3\widehat{i}-\widehat{j}+\widehat{k} \right)=1$ is
3x – y + z = 1 …..(1)
The direction cosines of the plane $\overset{\to }{\mathop{r}}\,\cdot \left( \widehat{i}+4\widehat{j}-2\widehat{k} \right)=2$ are (1, 4, -2) and perpendicular distance from origin is 2. Therefore the Cartesian equation of planes $\overset{\to }{\mathop{r}}\,\cdot \left( \widehat{i}+4\widehat{j}-2\widehat{k} \right)=2$ is
x + 4y - 2z = 2 …..(2)
Put z = t in equation (1) and equation (2), we get
3x – y + t = 1
x + 4y - 2z = 2
$\Rightarrow$3x – y = 1 - t …..(3)
And x + 4y = 2 + 2t…..(4)
Multiplying equation (4) by 3, we get
3x + 12y = 6 + 6t…..(5)
Subtracting equation (3) from equation (5), we get
13y = 5 + 7t
$y=\dfrac{\text{ }5\text{ }+\text{ }7t}{13}$
Using the value of y in equation (3), we get
$3x\text{ }\text{- }\dfrac{\text{ }5\text{ }+\text{ }7t}{13}\text{ }=\text{ }1\text{ }-\text{ }t$
By adding $\dfrac{\text{ }5\text{ }+\text{ }7t}{13}$ on both sides, we get
$3x\text{ }\text{ -}\dfrac{\text{ }5\text{ }+\text{ }7t}{13}\text{+}\dfrac{\text{ }5\text{ }+\text{ }7t}{13}\text{ }=\text{ }1\text{ }-\text{ }t+\dfrac{\text{ }5\text{ }+\text{ }7t}{13}$
$3x\text{ }=\text{ }1\text{ }-\text{ }t+\dfrac{\text{ }5\text{ }+\text{ }7t}{13}$
By cross multiplication, we get
$3x\text{ }=\dfrac{\text{13-13t+ }5\text{ }+\text{ }7t}{13}$
$3x\text{ }=\dfrac{\text{1}8-6t}{13}$
By dividing 3 on both sides, we get
$x\text{ }=\dfrac{6-2t}{13}$
Hence parametric equations of line are
$x\text{ }=\dfrac{6-2t}{13},\text{ }y=\dfrac{\text{ }5\text{ }+\text{ }7t}{13},\text{ }z=t$
By eliminating t we get the cartesian form of the line as follows
$\dfrac{x-\dfrac{6}{13}}{2}=\dfrac{y-\dfrac{5}{13}}{7}=\dfrac{z}{-13}$ .

So, the correct answer is “Option A”.

Note: The different forms of equation of line is given below,\\

1. Vector form: let $\overset{\to }{\mathop{a}}\,$ be the position vector of point A with respect to origin. Let L be a line which passes through point A and is parallel to a given vector$\overset{\to }{\mathop{b}}\,$. Let $\overset{\to }{\mathop{r}}\,$ be the position vector of the arbitrary point P on the line. Then the vector form of equation of line is
   $\overset{\to }{\mathop{r}}\,=\overset{\to }{\mathop{a}}\,+\lambda \overset{\to }{\mathop{b}}\,$
2. Let the coordinates of given point A be $\left( {{x}_{1}},\text{ }{{y}_{1}},\text{ }{{z}_{2}} \right)$and the direction ratio of the lines be a, b, c. Then the parametric equation of line is
   $x\text{ }=\text{ }{{x}_{1}}+\text{ }ka,\text{ }y\text{ }=\text{ }{{y}_{1}}+\text{ }kb\text{ }and\text{ }z\text{ }=\text{ }{{z}_{2}}+\text{ }kc.$
   The by eliminating k, we get cartesian form of line
   $\dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}$