Respuesta :
Answer:
The values of x that makes these vectors orthogonal are x = 2 and x = 4.
Step-by-step explanation:
Orthogonal vectors
Suppose we have two vectors:
[tex]v_{1} = (a,b,c)[/tex]
[tex]v_{2} = (d,e,f)[/tex]
Their dot product is:
[tex](a,b,c).(d,e,f) = ad + be + cf[/tex]
They are ortogonal is their dot product is 0.
Solving quadratic equations:
To solve this problem, we are going to need tosolve a quadratic equation.
Given a second order polynomial expressed by the following equation:
[tex]ax^{2} + bx + c, a\neq0[/tex].
This polynomial has roots [tex]x_{1}, x_{2}[/tex] such that [tex]ax^{2} + bx + c = (x - x_{1})*(x - x_{2})[/tex], given by the following formulas:
[tex]x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}[/tex]
[tex]x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}[/tex]
[tex]\bigtriangleup = b^{2} - 4a[/tex]
Find all values of x such that (4, x, −6) and (2, x, x) are orthogonal.
[tex](4,x,-6)(2,x,x) = 8 + x^{2} - 6x[/tex]
These vectors are going to be orthogonal if:
[tex]x^{2} -6x + 8 = 0[/tex]
This is a quadratic equation, in which [tex]a = 1, b = -6, c = 8[/tex]. So
[tex]\bigtriangleup = 6^{2} - 4*1*8 = 4[/tex]
[tex]x_{1} = \frac{-(-6) + \sqrt{4}}{2} = 4[/tex]
[tex]x_{2} = \frac{-(-6) - \sqrt{4}}{2} = 2[/tex]
The values of x that makes these vectors orthogonal are x = 2 and x = 4.
