Find the values of x such that the angle between the vectors (2, 1, -1) and (1, x, 0) is 40.

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

Angle

Dot product

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:

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

Cosine between two vectors

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:

$x = 0.804$