Answer :
[tex](3,2,x)[/tex] and [tex](2x,4,x)[/tex] are orthogonal if their dot product is 0:
[tex](3,2,x)\cdot(2x,4,x)=6x+8+x^2=(x+3)^2-1=0\implies x=-3\pm1[/tex]
[tex]\implies x=-4\text{ or }x=-2[/tex]
Applying the dot product, it is found that the values of x are x = -2 and x = -4.
What are orthogonal vectors?
Two vectors are orthogonal if their dot product is zero.
The dot product between two vectors is the sum of the multiplications of each respective term.
In this problem, the dot product is:
[tex](3, 2, x).(2x, 4,x) = 6x + 8 + x^2[/tex]
Hence:
[tex]x^2 + 6x + 8 = 0[/tex]
Which is a quadratic equation with coefficients [tex]a = 1, b = 6, c = 8[/tex].
Then:
[tex]\Delta = b^2 - 4ac = 6^2 - 4(1)(8) = 4[/tex]
[tex]x_1 = \frac{-b + \sqrt{\Delta}}{2a} = \frac{-6 + 2}{2} = -2[/tex]
[tex]x_2 = \frac{-b - \sqrt{\Delta}}{2a} = \frac{-6 - 2}{2} = -4[/tex]
The values of x are x = -2 and x = -4.
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