The question aims to find the value of an **unknown** variable given in **3D vector coordinates** and the **angle** between those **vectors.**

The question depends on the **dot product** of two **3D vectors** to calculate the **angle** between those vectors. As the **angle** is already given, we can use the **equation** to calculate the unknown coordinate of the vector. It also depends on the **magnitude** of the **vector** as we need the **magnitude** of the vector to calculate the **cosine** between **two** **vectors.** The formula for **magnitude** of any vector is given as:

\[ |\ \overrightarrow{a}\ | = \sqrt{ {a_x}^2 + {a_y}^2 + {a_z}^2 } \]

## Expert Answer

The given vectors **A** and **B **are:

\[ \overrightarrow{A} = < 2, -1, 1 > \]

\[ \overrightarrow{B} = < 1, x, 0 > \]

To find the value of **unknown value ‘x’,** we can take the **dot product** of these **two vectors** as we already know the **angle** between those **vectors.** The equation for **dot product** of these vectors is given as:

\[ < 2, -1, 1 > . < 1, x, 0 > = |A| |B| \cos \theta \]

\[ (2)(1) + (-1)(x) + (1)(0) = \sqrt{ 2^2 + (-1)^2 + 1^2 } \sqrt{ 1^2 + x^2 + 0^2 } \cos (40) \]

\[ 2\ -\ x + 0 = \sqrt{ 4 + 1 + 1 } \sqrt{ 1 + x^2 } \times 0.766 \]

\[ 2\ -\ x = \sqrt{6} \sqrt{1 + x^2} \times 0.766 \]

**Dividing 0.766** on both sides:

\[ \dfrac{ 2\ -\ x }{ 0.766 } = \sqrt{ 6 + 6x^2 } \]

\[ – 1.31x + 2.61 = \sqrt { 6 + 6x^2 } \]

**Taking square** on both sides:

\[ (- 1.31x + 2.61)^2 = 6 + 6x^2 \]

\[ 1.7x^2\ -\ 6.82x + 6.82 = 6x^2 + 6 \]

\[ 4.3x^2 + 6.8x\ -\ 0.82 = 0 \]

Using the **quadratic formula** to find the value of **‘x’,** we get:

\[ x = [ 0.11, -1.69 ] \]

## Numerical Result

The value of **unknown coordinate** in the **vector** is calculated to be:

\[ x = [ 0.11, -1.69 ] \]

The **angle** between **two vectors** will be $40^{\circ}$ for both values of **x**.

## Example

Find the **unknown value** of the vector given below such that the **angle** between those vectors is **60.**

\[ a(-1, 0, 1) \]

\[ b(x, 0, 3) \]

Taking the **dot product** of these vectors as we already have the **angle** between them. The **dot product** is given as:

\[ < -1, 0, 1 > . < x, 0, 3 > = |a| |b| \cos \theta \]

\[ -x + 0 + 3 = \sqrt{ 1 + 0 + 1 } \sqrt{ x^2 + 0 + 9 } \cos (60) \]

\[ -x + 3 = \sqrt{2} \sqrt{ x^2 + 9 } \dfrac{1}{2} \]

\[ -x + 3 = \sqrt{ x^2 + 9 } \dfrac{ 1 }{ \sqrt{2} } \]

\[ -x + 3 = 0.707 \sqrt{x^2 + 9} \]

\[ -1.41x + 4.24 = \sqrt{x^2 + 9} \]

\[ 1.99x^2\ -\ 11.99x + 17.99 = x^2 + 9 \]

\[ -0.999x^2 + 11.99x\ -\ 8.99 = 0 \]

Using the **quadratic formula** to find the value of **‘x’,** we get:

\[ x = 0.804 \]