Will medal and fan if you help me solve this beast The plane through (1,2,−1) that is perpendicular to the line of intersection of the planes 2x+y+z=2 and x + 2y + z = 3.
looks fine
? Do you know how to solve it ? @IrishBoy123 I am not sure what you mean by "it looks fine"
what i mean is that i did what the did and got the same answers. do you get something different?
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@IrishBoy123 If you are refering to the previews question I posted.... I think he replaced x rong... because he sets x to 1.... but x vanishes completely ... that's for the previews question...
sets x to 0 ... x vanishes ... sets x to 1 .... x vanishes again ...
nah, look how the -2 becomes a 2 it's done in one step
It's a two step solution. 1. compute the vector \(\vec v_3 = \vec v_1 \times\vec v_2 = (a,b,c)\) as @mathmate illustrated -> this is the normal vector of the plane to find 2. use the fact that \((1,2,-1)\) belongs to the plane to find the \(d\) in the equation \[ax+by+cz+d=0.\]
Do you find \(\vec v_3 = (-1,-1,3)\) ?
@Christos
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