Answered: "Find the values of x such that the..."

High School
Geometry
Trigonometry
Right triangles and trigonometry

tearstreakdl

Answered question

2021-12-14

Find the values of x such that the angle between the vectors (2, 1, -1), and (1, x, 0) is

45^{\circ}

Answer & Explanation

Hector Roberts

Beginner2021-12-15Added 31 answers

Theorem 3 from this section states that:
$a \cdot b = m i d a p a r a l \leq l b m i d \cos θ$
where \theta is the angle between the vectors a,b
$⟨ 2, 1, - 1 ⟩ \cdot ⟨ 1, x, 0 ⟩ = \sqrt{2^{2} + 1^{2} + {(- 1)}^{2}} \sqrt{1^{2} + x^{2} + 0^{{}} {\cos 45}^{\circ}$
Remember that:
$⟨ a_{1}, b_{1}, c_{1} ⟩ \cdot ⟨ a_{2}, b_{2}, c_{2} ⟩ = a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}$
This is due to the fact that the unit vectors I j, and k have a dot product of 1 with themselves and a dot product of 0 with the other two.
$(2) (1) + (1) (x) + (- 1) (0) = \sqrt{4 + 1 + 1} \sqrt{1 + x^{2} + 0} \frac{1}{\sqrt{2}}$
$2 + x + 0 = \sqrt{6} \sqrt{1 + x^{2}} \frac{1}{\sqrt{2}}$
$2 + x = \sqrt{3} \sqrt{1 + x^{2}}$
Square both sides
${(2 + x)}^{2} = 3 (1 + x^{2})$
$4 + 4 x + x^{2} = 3 + 3 x^{2}$
The quadratic equation should now be written in standard form.
$2 x^{2} - 4 x - 1 = 0$
Using the quadratic formula, we can write
$x = \frac{- (- 4) \pm \sqrt{{(- 4)}^{2} - 4 (2) (- 1)}}{2 (2)} = \frac{4 \pm \sqrt{24}}{2 (2)} = \frac{2 \pm \sqrt{6}}{2}$

sukljama2

Beginner2021-12-16Added 32 answers

Step 1
Consider the vectors:

v = ⟨ 2, 1, - 1 ⟩

w = ⟨ 1, x, 0 ⟩

θ = 45^{\circ}

\vec{v} 2 i + j - k

\vec{w} = i + x j

\vec{v} \cdot \vec{w} = (2 i + j - k) \cdot (i + k)

= (2) (1) + (1) (x) + (- 1) (0)

= 2 + x

Step 2
Find the magnitude each vector:

m i d \vec{v} m i d = \sqrt{2^{2} + 1^{2} + {(1)}^{2}}

= \sqrt{4 + 1 + 1}

= \sqrt{6}

m i d \vec{w} m i d = \sqrt{1^{2} + {(x)}^{2} + {(0)}^{2}}

= \sqrt{1 + x^{2} + 0}

= \sqrt{1 + x^{2}}

Step 3 Find the cosine angle between two vectors:

\vec{v} \cdot \vec{w} = m i d \vec{v} p a r a l \leq l \vec{w} m i d \cos θ

2 + x = (\sqrt{6}) (\sqrt{1 + x^{2}}} {\cos 45}^{\circ}

2 + x = (\sqrt{6}) (\sqrt{1 + x^{2})} (\begin{matrix} \frac{1}{\sqrt{2}} \end{matrix})

2 + x = (\sqrt{6}) (\sqrt{1 + x^{2})} (\begin{matrix} \frac{\sqrt{2}}{2} \end{matrix})

2 + x = \frac{\sqrt{12}}{2} (\sqrt{1 + x^{2}}}

\frac{\sqrt{12}}{2} (2 + x) = \sqrt{1 + x^{2}}

\frac{4}{12} {(2 + x)}^{2} = 1 + x^{2}

\frac{4}{12} (4 + x^{2} + 4 x) = 1 + x^{2}

\frac{1}{3} (4 + x^{2} + 4 x) = 1 + x^{2}

This is helpful 0 Flag

alenahelenash

Expert2023-06-12Added 556 answers

Result:

x = \frac{1}{2} \pm \frac{\sqrt{2}}{2} .

Solution:
The dot product of

𝐯_{1}

and

𝐯_{2}

is given by:

𝐯_{1} \cdot 𝐯_{2} = 2 (1) + 1 (x) + (- 1) (0) = 2 + x .

The magnitudes of

𝐯_{1}

and

𝐯_{2}

are:

| 𝐯_{1} | = \sqrt{2^{2} + 1^{2} + (- 1)^{2}} = \sqrt{6},

| 𝐯_{2} | = \sqrt{1^{2} + x^{2} + 0^{2}} = \sqrt{1 + x^{2}} .

Now, using the dot product formula and the fact that the angle between the vectors is 45 degrees (which corresponds to

\cos (45^{\circ}) = \frac{1}{\sqrt{2}}

), we can set up the equation:

𝐯_{1} \cdot 𝐯_{2} = | 𝐯_{1} | \cdot | 𝐯_{2} | \cdot \cos (45^{\circ}) .

Substituting the values, we have:

2 + x = \sqrt{6} \cdot \sqrt{1 + x^{2}} \cdot \frac{1}{\sqrt{2}} .

Simplifying the equation, we get:

2 + x = \sqrt{3 (1 + x^{2})} .

Squaring both sides, we obtain:

(2 + x)^{2} = 3 (1 + x^{2}) .

Expanding and simplifying:

4 + 4 x + x^{2} = 3 + 3 x^{2} .

Rearranging the terms:

2 x^{2} - 4 x + 1 = 0 .

To solve this quadratic equation, we can use the quadratic formula:

x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a},

where

a = 2

b = - 4

, and

c = 1

.
Evaluating the formula, we get:

x = \frac{- (- 4) \pm \sqrt{(- 4)^{2} - 4 (2) (1)}}{2 (2)} .

Simplifying further:

x = \frac{4 \pm \sqrt{16 - 8}}{4} .

x = \frac{4 \pm \sqrt{8}}{4} .

x = \frac{4 \pm 2 \sqrt{2}}{4} .

Finally, we can write the solutions as:

x = \frac{1}{2} \pm \frac{\sqrt{2}}{2} .

karton

Expert2023-06-12Added 613 answers

To find the values of x such that the angle between the vectors

𝐯 = (2, 1, - 1)

and

𝐰 = (1, x, 0)

is 45 degrees, we can use the formula for the dot product of two vectors:

𝐯 \cdot 𝐰 = | 𝐯 | \cdot | 𝐰 | \cdot \cos (θ),

where

θ

is the angle between the vectors,

| 𝐯 |

and

| 𝐰 |

are the magnitudes of the vectors, and

\cdot

denotes the dot product.
In this case,

| 𝐯 | = \sqrt{2^{2} + 1^{2} + (- 1)^{2}} = \sqrt{6}

and

| 𝐰 | = \sqrt{1^{2} + x^{2} + 0^{2}} = \sqrt{1 + x^{2}}

.
The dot product of

𝐯

and

𝐰

is given by

𝐯 \cdot 𝐰 = 2 \cdot 1 + 1 \cdot x + (- 1) \cdot 0 = 2 + x

.
Substituting these values into the dot product formula, we have:

(2 + x) = \sqrt{6} \cdot \sqrt{1 + x^{2}} \cdot \cos (45^{\circ}) .

Since

\cos (45^{\circ}) = \frac{\sqrt{2}}{2}

, we can simplify the equation as:

(2 + x) = \frac{\sqrt{6}}{2} \cdot \sqrt{1 + x^{2}} .

To solve for x, we can square both sides of the equation:

(2 + x)^{2} = \frac{6}{4} \cdot (1 + x^{2}) .

Expanding and simplifying:

4 + 4 x + x^{2} = \frac{3}{2} + \frac{3}{2} x^{2} .

Rearranging terms:

\frac{1}{2} x^{2} - 4 x + \frac{5}{2} = 0 .

To solve this quadratic equation, we can use the quadratic formula:

x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} .

For our equation,

a = \frac{1}{2}

b = - 4

, and

c = \frac{5}{2}

. Substituting these values into the quadratic formula, we have:

x = \frac{- (- 4) \pm \sqrt{(- 4)^{2} - 4 \cdot \frac{1}{2} \cdot \frac{5}{2}}}{2 \cdot \frac{1}{2}} .

Simplifying further:

x = \frac{4 \pm \sqrt{16 - 10}}{1} .

x = \frac{4 \pm \sqrt{6}}{1} .

Therefore, the solutions for x are:

x = 4 + \sqrt{6}

and

x = 4 - \sqrt{6} .

These are the values of x for which the angle between the vectors (2, 1, -1) and (1, x, 0) is 45 degrees.

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