Question

How do you find the angle between the vectors $v = i + j$, $w = 3i - j$?

Answer 1

Here the given question has two vectors, we have to find the angle between those two vectors. To find the angle $\theta$ between two vectors, use the formula for finding that angle's cosine. i.e., $cos\theta = \dfrac{{\left( {\overrightarrow u \cdot \overrightarrow v } \right)}}{{\left( {\left\| {\overrightarrow u } \right\|\,\left\| {\overrightarrow v } \right\|} \right)}}$ where $\left\| {\overrightarrow u } \right\|$ and $\left\| {\overrightarrow v } \right\|$ means “the length of vector $\overrightarrow u$ and $\overrightarrow v$” respectively and $\overrightarrow u \cdot \overrightarrow v$ is the dot product (scalar product) of the two vectors, on simplifying the formula we get the required angle.

Complete step by step solution:
a vector is any object that has a definable length, known as magnitude, and direction. Since vectors are not the same as standard lines or shapes, you’ll need to use some special formulas to find angles between them.
To find the angle θ between two vectors, start with the formula for finding that angle's cosine.
i.e., $cos\theta = \dfrac{{\left( {\overrightarrow u \cdot \overrightarrow v } \right)}}{{\left( {\left\| {\overrightarrow u } \right\|\,\left\| {\overrightarrow v } \right\|} \right)}}$
$\left\| {\overrightarrow u } \right\|$ and $\left\| {\overrightarrow v } \right\|$ means “the length of vector $\overrightarrow u$ and $\overrightarrow v$” respectively
$\overrightarrow u \cdot \overrightarrow v$ is the dot product (scalar product) of the two vectors
Consider the given Two vectors
$v = i + j$ and $w = 3i - j$
Calculate the length of each vector. Picture a right triangle drawn from the vector's x-component, its y-component, and the vector itself. The vector forms the hypotenuse of the triangle, so to find its length we use the Pythagorean theorem.
Therefore, for a two-dimensional vector, $\left\| {\overrightarrow u } \right\| = \sqrt {u_1^2 + u_2^2}$.
Now, the length of vector $\overrightarrow v$ and $\overrightarrow w$ is
$\Rightarrow \,\,\,\,\left\| {\overrightarrow v } \right\| = \sqrt {v_1^2 + v_2^2}$ and $\Rightarrow \,\,\,\,\left\| {\overrightarrow w } \right\| = \sqrt {w_1^2 + w_2^2}$
$\Rightarrow \,\,\,\,\left\| {\overrightarrow v } \right\| = \sqrt {{1^2} + {1^2}}$ and $\Rightarrow \,\,\,\,\left\| {\overrightarrow w } \right\| = \sqrt {{3^2} + {{\left( { - 1} \right)}^2}}$
$\Rightarrow \,\,\,\,\left\| {\overrightarrow v } \right\| = \sqrt {1 + 1}$ and $\Rightarrow \,\,\,\,\left\| {\overrightarrow w } \right\| = \sqrt {9 + 1}$
$\Rightarrow \,\,\,\,\left\| {\overrightarrow v } \right\| = \sqrt 2$ and $\Rightarrow \,\,\,\,\left\| {\overrightarrow w } \right\| = \sqrt {10}$
Next, to calculate the dot product in terms of the vectors' components, multiply the components in each direction together i.e., $\overrightarrow u \cdot \overrightarrow v = {u_1}{v_1} + {u_2}{v_2}$
Now, the dot product of vector $\overrightarrow v \cdot \overrightarrow w$ is
$\Rightarrow \,\,\,\,\overrightarrow v \cdot \overrightarrow w = 1 \cdot 3 + 1 \cdot \left( { - 1} \right)$
$\Rightarrow \,\,\,\,\overrightarrow v \cdot \overrightarrow w = 3 - 1$
$\Rightarrow \,\,\,\,\overrightarrow v \cdot \overrightarrow w = 2$
Substitute length and dot product of vector $\overrightarrow v \cdot \overrightarrow w$ in cosine formula, then
$\Rightarrow \,\,\,\,cos\theta = \dfrac{2}{{\left( {\sqrt 2 \cdot \sqrt {10} } \right)}}$
$\Rightarrow \,\,\,\,cos\theta = \dfrac{2}{{\left( {\sqrt {10 \times 2} } \right)}}$
$\Rightarrow \,\,\,\,cos\theta = \dfrac{2}{{\left( {\sqrt {20} } \right)}}$
$\Rightarrow \,\,\,\,cos\theta = \dfrac{2}{{\left( {\sqrt {4 \times 5} } \right)}}$
$\Rightarrow \,\,\,\,cos\theta = \dfrac{2}{{\left( {2\sqrt 5 } \right)}}$
$\Rightarrow \,\,\,\,cos\theta = \dfrac{1}{{\sqrt 5 }}$
Take inverse cosine function both side, then
$\Rightarrow \,\,\,{\cos ^{ - 1}}\left( {cos\theta } \right) = {\cos ^{ - 1}}\left( {\dfrac{1}{{\sqrt 5 }}} \right)$
AS we know $x{x^{ - 1}} = 1$, then
$\Rightarrow \,\,\theta = {\cos ^{ - 1}}\left( {\dfrac{1}{{\sqrt 5 }}} \right)$

Hence the angle between the vectors $v = i + j$, $w = 3i - j$ is $\theta = {\cos ^{ - 1}}\left( {\dfrac{1}{{\sqrt 5 }}} \right)$.

Note: The angle between the vectors are defined by the angle of the cosine. In the vectors while multiplying the two vectors we have a dot product and the cross product. While the dot product is the same as the multiplication but the cross product of vectors is determined by the concept of determinant.