00:01
In the given question, we have to find a set of scalar parametric equation for the line formed by the two intersection plane, which is given as 3 times x plus 3 times y plus 2 times z plus 2 is equal to 0 and 2 times x minus 3 times y plus 2 times x minus 2 is equal to 0.
00:54
Now let we will consider that n1 vector as 3 .3 .2 and the vector n2 as 2.
01:10
2 comma negative 3 .2 be the normal vectors to the plane.
01:25
Now we will calculate the unit vector n were given as n1 cross n2.
01:34
So we have the product of 2 vector cross product 3 cross 3 cross 2 cross negative 3 cross 2 we simplifying gives us 6 plus 6 comma 4 minus 6 then we have minus 9 minus 6 so we get this is equals to 12 well, negative 2, negative 15.
02:09
We calculate these product n1 cross n2 by taking the determinant of i, j, k, then the vector 332 to negative 3 and 2.
02:25
Simplifying it, we get the unit vector n is equal to 4 if we divide this by 3, negative 2, negative 2.
02:35
Now to find the point of intersection of the planes in the yz plane we will choose x is equal to 0...