We need to calculate the magnitude, direction, and projections of vectors using unit vector notation and dot products.
Concept: Unit vector
The magnitude of a vector r = x i ^ + y j ^ \mathbf{r} = x\hat{\mathbf{i}} + y\hat{\mathbf{j}} r = x i ^ + y j ^
∣ r ∣ = x 2 + y 2 |\mathbf{r}| = \sqrt{x^2 + y^2} ∣ r ∣ = x 2 + y 2 Its direction angle is:
θ = tan − 1 ( y x ) \theta = \tan^{-1}\left(\frac{y}{x}\right) θ = tan − 1 ( x y ) The component (scalar projection) of vector A \mathbf{A} A B \mathbf{B} B
A B = A ⋅ B ∣ B ∣ A_B = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{B}|} A B = ∣ B ∣ A ⋅ B Given:
Unit vectors: i ^ \hat{\mathbf{i}} i ^ j ^ \hat{\mathbf{j}} j ^
Vector 1: u = i ^ + j ^ \mathbf{u} = \hat{\mathbf{i}} + \hat{\mathbf{j}} u = i ^ + j ^
Vector 2: v = i ^ − j ^ \mathbf{v} = \hat{\mathbf{i}} - \hat{\mathbf{j}} v = i ^ − j ^
Target vector: A = 2 i ^ + 3 j ^ \mathbf{A} = 2\hat{\mathbf{i}} + 3\hat{\mathbf{j}} A = 2 i ^ + 3 j ^
Solving:
Magnitude and direction of u = i ^ + j ^ \mathbf{u} = \hat{\mathbf{i}} + \hat{\mathbf{j}} u = i ^ + j ^
∣ u ∣ = 1 2 + 1 2 = 2 \begin{aligned}
|\mathbf{u}| &= \sqrt{1^2 + 1^2} \\
&= \sqrt{2}
\end{aligned} ∣ u ∣ = 1 2 + 1 2 = 2 θ u = tan − 1 ( 1 1 ) = 45 ∘ \begin{aligned}
\theta_u &= \tan^{-1}\left(\frac{1}{1}\right) \\
&= 45^\circ
\end{aligned} θ u = tan − 1 ( 1 1 ) = 4 5 ∘ The vector makes an angle of 45 ∘ 45^\circ 4 5 ∘
Magnitude and direction of v = i ^ − j ^ \mathbf{v} = \hat{\mathbf{i}} - \hat{\mathbf{j}} v = i ^ − j ^
∣ v ∣ = 1 2 + ( − 1 ) 2 = 2 \begin{aligned}
|\mathbf{v}| &= \sqrt{1^2 + (-1)^2} \\
&= \sqrt{2}
\end{aligned} ∣ v ∣ = 1 2 + ( − 1 ) 2 = 2 θ v = tan − 1 ( − 1 1 ) = − 45 ∘ (or 315 ∘ ) \begin{aligned}
\theta_v &= \tan^{-1}\left(\frac{-1}{1}\right) \\
&= -45^\circ \text{ (or } 315^\circ)
\end{aligned} θ v = tan − 1 ( 1 − 1 ) = − 4 5 ∘ (or 31 5 ∘ ) The vector makes an angle of − 45 ∘ -45^\circ − 4 5 ∘
Components of A = 2 i ^ + 3 j ^ \mathbf{A} = 2\hat{\mathbf{i}} + 3\hat{\mathbf{j}} A = 2 i ^ + 3 j ^
Component along u = i ^ + j ^ \mathbf{u} = \hat{\mathbf{i}} + \hat{\mathbf{j}} u = i ^ + j ^
A u = A ⋅ u ∣ u ∣ = ( 2 ) ( 1 ) + ( 3 ) ( 1 ) 2 = 5 2 \begin{aligned}
A_u &= \frac{\mathbf{A} \cdot \mathbf{u}}{|\mathbf{u}|} \\
&= \frac{(2)(1) + (3)(1)}{\sqrt{2}} \\
&= \frac{5}{\sqrt{2}}
\end{aligned} A u = ∣ u ∣ A ⋅ u = 2 ( 2 ) ( 1 ) + ( 3 ) ( 1 ) = 2 5
Component along v = i ^ − j ^ \mathbf{v} = \hat{\mathbf{i}} - \hat{\mathbf{j}} v = i ^ − j ^
A v = A ⋅ v ∣ v ∣ = ( 2 ) ( 1 ) + ( 3 ) ( − 1 ) 2 = − 1 2 \begin{aligned}
A_v &= \frac{\mathbf{A} \cdot \mathbf{v}}{|\mathbf{v}|} \\
&= \frac{(2)(1) + (3)(-1)}{\sqrt{2}} \\
&= -\frac{1}{\sqrt{2}}
\end{aligned} A v = ∣ v ∣ A ⋅ v = 2 ( 2 ) ( 1 ) + ( 3 ) ( − 1 ) = − 2 1 Answer
Magnitude of both i ^ + j ^ \hat{\mathbf{i}} + \hat{\mathbf{j}} i ^ + j ^ i ^ − j ^ \hat{\mathbf{i}} - \hat{\mathbf{j}} i ^ − j ^
2 \sqrt{2} 2 Directions:
45 ∘ and − 45 ∘ 45^\circ \quad \text{and} \quad -45^\circ 4 5 ∘ and − 4 5 ∘ Components of A \mathbf{A} A
5 2 and − 1 2 \frac{5}{\sqrt{2}} \quad \text{and} \quad -\frac{1}{\sqrt{2}} 2 5 and − 2 1