Solve this quadratic equation. x2 + 5x + 3 = 0
1 answer:
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Vectors:
i = [1 0 0]
j = [0 1 0]
k = [0 0 1]
i + j = [1 1 0]
i + k = [1 0 1]
We want a vector which is orthogonal to both i + j and i + k, so the escalar procut between our vector "v" and i + j, i + k is equal to 0
Let's say v = [x y z]
v . (i + j) = [x y z] . [1 1 0] = x.1 + y.1 + z.0 = 0 => x + y = 0
v . (i + k) = [x y z] . [1 0 1] = x.1 + y.0 + z.1 = 0 => x + z = 0
So, we can say:
x + y = 0 => x = -y
x + z = 0 => x = -z
If x = -y:
x + z = 0
-y + z = 0
z = y = a and x = -a
We know it's a unit vector, so:
So,
y = z = a = and x = -a =
v = [ ]