Question

Find the angle between the $\vec F = (4\hat i + 3\hat j)$ units, and displacement $\vec d = (4\hat j - 3\hat k)$ unit. Find the projection of $\vec F$ on $\vec d$ .

Answer 1

Hint : The rule of the dot product can be used to find the angle between two vectors. To find the dot product, we need to multiply only the components that are the same, and add all components together.

Formula used: In this solution we will be using the following formula;
$\vec A \cdot \vec B = \left| {\vec A} \right|\left| {\vec B} \right|\cos \theta$ where $\vec A \cdot \vec B$ is the dot product of vector $\vec A$ and $\vec B$ , $\left| {\vec A} \right|$ is the magnitude of $\vec A$ and $\left| {\vec B} \right|$ is the magnitude of $\vec B$ , and $\theta$ is the angle between them.
$\vec A \cdot \vec B = {A_x}{B_x} + {A_y}{B_y} + {A_z}{B_z}$ where the subscript x, y, z signifies the component of the vector. $a = \vec A \cdot \dfrac{{\vec B}}{{\left| {\vec B} \right|}}$ where $a$ is the projection of vectors $\vec A$ on $\vec B$ .

Complete step by step answer:
The angle between two vectors $\vec F = (4\hat i + 3\hat j)$ units and $\vec d = (4\hat j - 3\hat k)$ units are to be found.
The formula for dot product can be given as
$\vec A \cdot \vec B = \left| {\vec A} \right|\left| {\vec B} \right|\cos \theta$ where $\vec A \cdot \vec B$ is the dot product of vector $\vec A$ and $\vec B$ , $\left| {\vec A} \right|$ is the magnitude of $\vec A$ and $\left| {\vec B} \right|$ is the magnitude of $\vec B$ , and $\theta$ is the angle between them.
Hence, $\cos \theta = \dfrac{{\vec A \cdot \vec B}}{{\left| {\vec A} \right|\left| {\vec B} \right|}}$
Now to find dot product, we have
$\vec A \cdot \vec B = {A_x}{B_x} + {A_y}{B_y} + {A_z}{B_z}$ where the subscript x, y, z signifies the component of the vector.
Then $\vec F \cdot \vec d = (4\hat i + 3\hat j) \cdot (4\hat j - 3\hat k) = \left( {4 \times 4 + 3 \times - 3} \right)$
$\Rightarrow \vec F \cdot \vec d = 16 - 9 = 7{\text{square units}}$
Now the magnitude of a vector $\vec A$ is $\left| {\vec A} \right| = \sqrt {{A_x}^2 + {A_y}^2 + {A_z}^2}$
Then $\left| {\vec F} \right| = \sqrt {{4^2} + {3^2} + {0^2}} = 5$ and
$\left| {\vec d} \right| = \sqrt {{0^2} + {4^2} + {{\left( { - 3} \right)}^2}} = 5$
Hence, $\cos \theta = \dfrac{{\vec F \cdot \vec d}}{{\left| {\vec F} \right|\left| {\vec d} \right|}} = \dfrac{7}{{5 \times 5}} = \dfrac{7}{{25}}$
$\Rightarrow \theta = {\cos ^{ - 1}}\left( {\dfrac{7}{{25}}} \right) = 1.29rad$ or about $74^\circ$ (74 degrees)
The projection of the vector $\vec F$ on the vector $\vec d$ can be given as
$f = \vec F \cdot \dfrac{{\vec d}}{{\left| {\vec d} \right|}}$
This can be proven to be
$f = \left| {\vec F} \right|\cos \theta$
Hence, by substitution of the know values
$f = 5 \times \dfrac{7}{{25}}$
$\Rightarrow f = \dfrac{7}{5}$ .

Note:
For clarity, $f = \vec F \cdot \dfrac{{\vec d}}{{\left| {\vec d} \right|}}$ can be proven as follows.
$f = \vec F \cdot \dfrac{{\vec d}}{{\left| {\vec d} \right|}} = \dfrac{{\vec F \cdot \vec d}}{{\left| {\vec d} \right|}}$
If we multiply both sides by the magnitude of the vector $\vec F$ , we have
$f = \dfrac{{\left| {\vec F} \right|}}{{\left| {\vec F} \right|}}\dfrac{{\vec F \cdot \vec d}}{{\left| {\vec d} \right|}}$
This can be written as
$f = \left| {\vec F} \right|\dfrac{{\vec F \cdot \vec d}}{{\left| {\vec F} \right|\left| {\vec d} \right|}}$
From, the formula of product rule
$\cos \theta = \dfrac{{\vec F \cdot \vec d}}{{\left| {\vec F} \right|\left| {\vec d} \right|}}$
Hence, by replacing $\dfrac{{\vec F \cdot \vec d}}{{\left| {\vec F} \right|\left| {\vec d} \right|}}$ with $\cos \theta$
$f = \left| {\vec F} \right|\cos \theta$ .