The line of intersection of the planes \small \underset{r}{\rightarrow} .\left ( 3\hat{i} -\hat{j} +\hat{k}\right )= 1 and \small \underset{r}{\rightarrow} .\left ( \hat{i} +4 \hat{j} -2\hat{k}\right ) =2, is:
Option: 1 -2 \hat{i}+7 \hat{j}+13 \hat{k}
Option: 2 2 \hat{i}+7 \hat{j}-13 \hat{k}
Option: 3 -2 \hat{i}-7 \hat{j}+13 \hat{k}
Option: 4 -2 \hat{i}+7 \hat{j}+13 \hat{k}
 

Answers (1)

As we have learned

Equation of line as intersection of two planes -

Let the two intersecting planes be

ax+by+cz+d= 0 and 

a_{1}x+b_{1}y+c_{1}z+d_{1}= 0

then the parallel vector of line formed their intersection can be obtained by

\begin{vmatrix} \hat{i} &\hat{j} & \hat{k}\\ a&b &c \\ a_{1} & b_{1} & c_{1} \end{vmatrix}= A\hat{i}+B\hat{j}+C\hat{k}(assumed)

and points can be obtained by putting z= 0 and solving

ax+by+d= 0 and 

a_{1}x+b_{1}y+d_{1}= 0 say \alpha ,\beta

Now the equation will be

\frac{x-\alpha }{A}=\frac{y-\beta }{B}=\frac{z-0 }{C}

 

-

 

 3x-y + z = 1 \\ and \: \: x + 4y - 2z = 2

putting z = 0 

3x-y = 1    and   x + 4y = 2 

\Rightarrow 13 x = 6 \Rightarrow x = 6/13 \\ y = 5/3

vector \: \: along \: \: line = \begin{vmatrix} \hat i & \hat j &\hat k \\ 3 & -1 & 1\\ 1 & 4 &-2 \end{vmatrix} = -2 \hat i + 7 \hat j +13 \hat k

 

 

 

 

 

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