Question

3.1 vector addition and subtraction (15pts)
(problem 2.3.47 and more)
for vectors $vec{b}=-hat{i}-4hat{j}$ and $vec{a}=-3hat{i}-2hat{j}$, calculate magnitude and direction angles (in degrees) of:
(a) $vec{a}+vec{b}$
(b) $vec{a}-vec{b}$
(c) $vec{b}+vec{a}$
(d) $vec{b}-vec{a}$

Explanation:

Step1: Calculate the sum/difference vectors

(a) $\vec{A}+\vec{B}$

$\vec{A}+\vec{B}=(-3\hat{i}- 2\hat{j})+(-\hat{i}-4\hat{j})=(-3 - 1)\hat{i}+(-2-4)\hat{j}=-4\hat{i}-6\hat{j}$

(b) $\vec{A}-\vec{B}$

$\vec{A}-\vec{B}=(-3\hat{i}-2\hat{j})-(-\hat{i}-4\hat{j})=(-3+1)\hat{i}+(-2 + 4)\hat{j}=-2\hat{i}+2\hat{j}$

(c) $\vec{B}+\vec{A}$

Since vector addition is commutative, $\vec{B}+\vec{A}=\vec{A}+\vec{B}=-4\hat{i}-6\hat{j}$

(d) $\vec{B}-\vec{A}$

$\vec{B}-\vec{A}=(-\hat{i}-4\hat{j})-(-3\hat{i}-2\hat{j})=(-1 + 3)\hat{i}+(-4+2)\hat{j}=2\hat{i}-2\hat{j}$

Step2: Calculate the magnitude of the vectors

The magnitude of a vector $\vec{V}=x\hat{i}+y\hat{j}$ is given by $|\vec{V}|=\sqrt{x^{2}+y^{2}}$

(a) $|\vec{A}+\vec{B}|$

$|\vec{A}+\vec{B}|=\sqrt{(-4)^{2}+(-6)^{2}}=\sqrt{16 + 36}=\sqrt{52}=2\sqrt{13}\approx 7.21$

(b) $|\vec{A}-\vec{B}|$

$|\vec{A}-\vec{B}|=\sqrt{(-2)^{2}+2^{2}}=\sqrt{4 + 4}=\sqrt{8}=2\sqrt{2}\approx2.83$

(c) $|\vec{B}+\vec{A}|$

$|\vec{B}+\vec{A}| = |\vec{A}+\vec{B}|=2\sqrt{13}\approx 7.21$

(d) $|\vec{B}-\vec{A}|$

$|\vec{B}-\vec{A}|=\sqrt{2^{2}+(-2)^{2}}=\sqrt{4 + 4}=\sqrt{8}=2\sqrt{2}\approx2.83$

Step3: Calculate the direction - angle of the vectors

The direction - angle $\theta$ of a vector $\vec{V}=x\hat{i}+y\hat{j}$ is given by $\theta=\tan^{-1}(\frac{y}{x})$

(a) $\theta_{A + B}$

$\theta_{A + B}=\tan^{-1}(\frac{-6}{-4})=\tan^{-1}(1.5)$
Since the vector $-4\hat{i}-6\hat{j}$ is in the third - quadrant, $\theta_{A + B}=180^{\circ}+\tan^{-1}(1.5)\approx180^{\circ}+56.31^{\circ}=236.31^{\circ}$

(b) $\theta_{A - B}$

$\theta_{A - B}=\tan^{-1}(\frac{2}{-2})=\tan^{-1}(-1)$
Since the vector $-2\hat{i}+2\hat{j}$ is in the second - quadrant, $\theta_{A - B}=180^{\circ}+\tan^{-1}(-1)=135^{\circ}$

(c) $\theta_{B + A}$

$\theta_{B + A}=\theta_{A + B}\approx236.31^{\circ}$

(d) $\theta_{B - A}$

$\theta_{B - A}=\tan^{-1}(\frac{-2}{2})=\tan^{-1}(-1)$
Since the vector $2\hat{i}-2\hat{j}$ is in the fourth - quadrant, $\theta_{B - A}=360^{\circ}+\tan^{-1}(-1)=315^{\circ}$

Answer:

(a) Magnitude: $2\sqrt{13}\approx7.21$, Direction angle: $236.31^{\circ}$
(b) Magnitude: $2\sqrt{2}\approx2.83$, Direction angle: $135^{\circ}$
(c) Magnitude: $2\sqrt{13}\approx7.21$, Direction angle: $236.31^{\circ}$
(d) Magnitude: $2\sqrt{2}\approx2.83$, Direction angle: $315^{\circ}$