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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)

best_answer

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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