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Find values so that the vector has the specified angle

  • Thread starter Calpalned
  • Start date
  • #1
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Homework Statement


Find the values of x such that the angle between the vectors <2, 1, -1>, and <1, x, 0> is 45 degrees.

Homework Equations


<2, 1, -1> dot <1, x, 0> = 2 + x
|first vector| = (6)0.5
|2nd vector| = (1 + x2)1/2
Angle = cos-1(a dot b)/(|a||b|)
Cosine of 45 is the root of two divided by two.

The Attempt at a Solution


I set (2 + x) divided by 6^0.5 and (1 + x2)^0.5 equal to 1 divided by the root of 2.
I eventually get 2√2 + x√2 = √6√1 + x2
Am I on the right track?
 

Answers and Replies

  • #2
33,271
4,975

Homework Statement


Find the values of x such that the angle between the vectors <2, 1, -1>, and <1, x, 0> is 45 degrees.

Homework Equations


<2, 1, -1> dot <1, x, 0> = 2 + x
|first vector| = (6)0.5
|2nd vector| = (1 + x2)1/2
Angle = cos-1(a dot b)/(|a||b|)
Cosine of 45 is the root of two divided by two.

The Attempt at a Solution


I set (2 + x) divided by 6^0.5 and (1 + x2)^0.5 equal to 1 divided by the root of 2.
Or equivalently, ##\frac{2 + x}{\sqrt{6}\sqrt{1 + x^2}} = \frac{\sqrt{2}}{2}##
Calpalned said:
I eventually get 2√2 + x√2 = √6√1 + x2
I'm not sure about the above, although it might be right. I get a quadratic with all integer coefficients.
Can you show what you did going from your equation above to this one?
Calpalned said:
Am I on the right track?
 

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