Question

Find all values of x such that (4, x, −6) and (2, x, x) are orthogonal. (Enter your answers as a comma-separated list.)

Answer 1

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:

$v_{1} = (a,b,c)$

$v_{2} = (d,e,f)$

Their dot product is:

$(a,b,c).(d,e,f) = ad + be + cf$

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:

$ax^{2} + bx + c, a\neq0$.

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = (x - x_{1})*(x - x_{2})$, given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4a$

Find all values of x such that (4, x, −6) and (2, x, x) are orthogonal.

$(4,x,-6)(2,x,x) = 8 + x^{2} - 6x$

These vectors are going to be orthogonal if:

$x^{2} -6x + 8 = 0$

This is a quadratic equation, in which $a = 1, b = -6, c = 8$. So

$\bigtriangleup = 6^{2} - 4*1*8 = 4$

$x_{1} = \frac{-(-6) + \sqrt{4}}{2} = 4$

$x_{2} = \frac{-(-6) - \sqrt{4}}{2} = 2$

The values of x that makes these vectors orthogonal are x = 2 and x = 4.