Question

Need help with finding the hyperplane for a=<-2, 1 ,4>

Answer 1

like for a=<0, -3, 5, 7> I don't know why the answer is (1,0,0,0),(7,5,3,0),(0,7,0,3)

Answer 2

for your example isn't the answer the span of those 3 vectors

Answer 3

Okay how would you get the span of these 3 vectors then so frustrated I tried this so many times already

Answer 4

Am I also assume you are trying to find the hyperplane that goes through the origin?

Answer 5

yes

Answer 6

typically the hyperplane through the point P is the set of all vectors y such that

\[n\cdot(y-p)=0\]

where n is called the normal vector

Answer 7

so you need to find all the vectors y such that

\[a\cdot y=0\]

Answer 8

have you done the Gram Schmidt orthoginalization algorithm before?

Answer 9

Okay this is what I did I maade -2x_1 + x_2 + 4x_3 = 0

Answer 10

(-2x_1 + x_2 + 4x_3, x_2,x_3)=x_1(-2,0,0) + x_2(1,1,0) + x_3(4,0,1)

Answer 11

then v_1 =(-2 0 0 ) v_2=(1 1 0) v_3 =(4 0 1)

Answer 12

is that wrong? it looks wrong for some reason

Answer 13

Sorry I have to go....

it does look wrong...look into the algorithm I mentioned

Answer 14

try this ..pick any vector that is orthogonal to <-2, 1 ,4>

like <1,2,0>

now pick another that is orthogonal to <-2, 1 ,4>

<2,0,1>

<1,2,0> and <2,0,1> are clearly independent of each other

so the hyperplane is the span(<1,2,0>,<2,0,1>)