Question

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

Answer 1

Answer:

it's reseltunt vector

Answer 2

The equation of line intersection of the plane:

$\vec{r}=(\dfrac{6}{13}i-\dfrac{34}{13}j+0k)+t(-2i+7j+13k)$

Step-by-step explanation:

Equation of plane 1:

$r\cdot(3i-j+k)=1$

Normal to plane 1:

$n_1=3i-j+k$

Equation of plane 2:

$r\cdot (i+4j-2k)=2$

Normal to plane 2:

$n_2=i+4j-2k$

The parallel vector of the intersection line of both plane,

$\vec{b}=n_1\times n_2$

$\vec_b}=(3i-j+k)\times i+4j-2k$

$\vec_b}=-2i+7j+13k$

Now find one point of intersection of plane.

Put z=0 then 3x-y=1 and x+4y=2

Solve for x and y

$x=\dfrac{6}{13}\text{ and }y=-\dfrac{34}{13}$

Position vector, $\vec{a}=\dfrac{6}{13}i-\dfrac{34}{13}j+0k$

The equation of line intersection of the plane:

$\vec{r}=\vec{a}+\vec{b}t$

$\vec{r}=(\dfrac{6}{13}i-\dfrac{34}{13}j+0k)+t(-2i+7j+13k)$

#Learn more:

https://brainly.in/question/3618587