Question

Can you please answer all parts with explanations trying to figure out where i am going wrong.

Given two vectors A = 4.00î + 7.00ġ and B = 5.00 î – 2.00ĵ, (a) find the magnitude of each vector; (b) write an expres- sion for the vector difference A - B using unit vectors; (c) find the magnitude and direction of the vector difference A - B. (d) In a vector diagram show A, B, and A – B, and also show that your diagram agrees qualitatively with your answer in part (c). e-Calculate the scalar product of vectors A and B. f. Find the magnitude and direction of vector C = AXB. g- Find the angle between vectors A and B.

Answer 1

Ans:

In general, a vector is defined as

$\vec{A}=a_{1}\hat{i}+a_{2}\hat{j}$

.

Here ay ay are the magnitudes of vector A along r and y directions

Given:

$\vec{A}=4\hat{i}+7\hat{j},\: \: \vec{B}=5\hat{i}-2\hat{j}$

a) The magnitude of each vector

In general, the magnitude of a vector can be defined as,

$\mid \vec{A}\mid =\sqrt{\mid a_{1}\mid ^{2}+\mid a_{2}\mid ^{2}}$

$So \: for\: \vec{A}, \:\: a_{1}=4,\: a_{2}=7$

$\mid \vec{A}\mid =\sqrt{\mid 4\mid ^{2}+\mid 7\mid ^{2}}=\sqrt{75}=8.66$

$for \: \vec{B},\: \: a_{1}=5,\: \: a_{2}=-2$

$\mid \vec{B}\mid =\sqrt{\mid 5\mid ^{2}+\mid -2\mid ^{2}}=\sqrt{21}=5.38$

b) Vector difference:

$\vec{A}- \vec{B}=(a_{1}-b_{1})\hat{i}+(a_{2}-b_{2})\hat{j}\: \: Here \: \: (a_{1},a_{2})=(4,7),\: \: (b_{1},b_{2})=(5,-2)$

$\vec{A}- \vec{B}=(4-5)\hat{i}+(7-(-2))\hat{j}$

$\vec{A}- \vec{B}=(-1)\hat{i}+(9)\hat{j}$

c) The magnitude of the vector difference:

$\mid \vec{A}- \vec{B}\mid =\sqrt{\mid -1\mid^{2} +\mid 9\mid^{2} }=\sqrt{82}=9.05$

e) Calculate the scalar product of vectors :

$\vec{A}. \vec{B} =(a_{1}.b_{1})+(a_{2}.b_{2})=(4*5)+(7*-2)=20-14=6$

f) Find the magnitude and direction of the vector  $\vec{C}=\vec{A}\times \vec{B}$

$\vec{A}\times \vec{B} =\begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ 4&7 &0 \\ 5&-2 &0 \end{vmatrix}=\hat{i}(0)-\hat{j}(0)+\hat{k}(-8-35)=-43\hat{k}$

$Magnitude \: of\: \vec{C}=\sqrt{\mid -49\mid ^{2}}=49,\: \: Direction=along \: \: z\: direction$

g) Angle between vectors A and B:

$\vec{A}.\vec{B}=\mid A\mid .\mid B\mid \cos \Theta$

$\cos \Theta =\frac{\vec{A}.\vec{B}}{ \mid A\mid .\mid B\mid}$

$\cos \Theta =\frac{6}{8.66*5.38}=0.128$

$\Theta =82.64^{o}$