Question

Find the angle between u =the square root of 5i-8j and v =the square root of 5i+j.

Answer 1

The angle between vector
`\vec{u} = 5\, \vec{i} - 8\, \vec{j}` and
`\vec{v} = 5\, \vec{i} + \, \vec{j}` is approximately
`1.21` radians, which is equivalent to approximately
`69.3^\circ`.

Explanation:

The angle between two vectors can be found from the ratio between:

• their dot products, and
• the product of their lengths.

To be precise, if
`\theta` denotes the angle between
`\vec{u}` and
`\vec{v}` (assume that
`0^\circ \le \theta < 180^\circ` or equivalently
`0 \le \theta < \pi`,) then:

`\displaystyle \cos(\theta) = \frac{\vec{u} \cdot \vec{v}}\`.

Dot product of the two vectors

The first component of
`\vec{u}` is
`5` and the first component of
`\vec{v}` is also

The second component of
`\vec{u}` is
`(-8)` while the second component of
`\vec{v}` is
`1`. The product of these two second components is
`(-8) * 1= (-8)`.

The dot product of
`\vec{u}` and
`\vec{v}` will thus be:

`\begin{aligned} \vec{u} \cdot \vec{v} = 5 * 5 + (-8) *1 = 17 \end{aligned}`.

Lengths of the two vectors

Apply the Pythagorean Theorem to both
`\vec{u}` and
`\vec{v}`:

• 
  `\| u \| = √(5^2 + (-8)^2) = √(89)`.
• 
  `\| v \| = √(5^2 + 1^2) = √(26)`.

Angle between the two vectors

Let
`\theta` represent the angle between
`\vec{u}` and
`\vec{v}`. Apply the formula
`\displaystyle \cos(\theta) = \frac{\vec{u} \cdot \vec{v}} u \` to find the cosine of this angle:

`\begin{aligned} \cos(\theta)&= \frac{\vec{u} \cdot \vec{v}} v \ = (17)/(√(89)\cdot √(26))\end{aligned}`.

Since
`\theta` is the angle between two vectors, its value should be between
`0\; \rm radians` and
`\pi \; \rm radians` (
`0^\circ` and
`180^\circ`.) That is:
`0 \le \theta < \pi` and
`0^\circ \le \theta < 180^\circ`. Apply the arccosine function (the inverse of the cosine function) to find the value of
`\theta`:

`\displaystyle \cos^(-1)\left((17)/(√(89)\cdot √(26))\right) \approx 1.21 \;\rm radians \approx 69.3^\circ` .