The line of intersection of the planes r(3 i- j+ k)=1 and r( i+4 j-2 k)=2, is:

Question

The line of intersection of the planes $\vec r\cdot(3\hat i-\hat j+\hat k)=1$ and $\vec r\cdot(\hat i+4\hat j-2\hat k)=2$, is:

Accepted Answer

$$\overrightarrow n  = \overrightarrow {{n_1}}  \times \overrightarrow {{n_2}} $$

$$ \Rightarrow $$   $$\left| {\begin{pmatrix} {\widehat i} & {\widehat j} & {\widehat k}  \
   3 & { - 1} & 1  \
   1 & 4 & { - 2}  \

\end{pmatrix} } \right| = \widehat i\left( { - 2} \right) - \widehat j\left( { - 7} \right) + \widehat k\left( {13} \right)$$

$$\overrightarrow n  =  - 2\widehat i + 7\widehat j + 13\widehat k$$

Now,

$3x - y + z = 1$

$x + 4y - 2z = 2$

but   z $=$ 0 \& solving the given

x $=$ 6/13 \& y = 5/13

$$ \therefore $$   required equation of a line is

$${{x - 6/13} \over 2} = {{y - 5/13} \over { - 7}} = {z \over { - 13}}$$

Solutions · 1