Given three vectors A = 24i + 33j, B = 55i - 12j and C = 2i + 43j (a) Find the magnitude of each vector. (b) Write an expression for the vector difference A - C. (c) Find the magnitude and direction of the vector difference A-B. (d) In a vector diagram show vector A + B, and A - B, and also show that your diagram agrees qualitatively with your answer.

Respuesta :

Answer:

(a) , .  and .

(b)[tex]\vec A - \vec C=22 \hat i -10 \hat j[/tex].

(c)[tex]|\vec A - \vec B|=63.13[/tex] and the direction [tex]\theta =[/tex] 124.56°.

Explanation:

Given that,

,

and

[tex]\vec {C}=2 \hat i +43 \hat j[/tex]

(a) The magnitude of a vector is the square root of the sum of the square of all the components of the vector, i.e. for a ,.

So, the magnitude of the is

[tex]|\vec A|=\sqrt {24^2+ 33^2}[/tex]

[tex]\Rightarrow |\vec A|=\sqrt {1665}[/tex]

.

The magnitude of the is

[tex]|\vec B|=\sqrt {55^2+ (-12)^2}[/tex]

[tex]\Rightarrow |\vec B|=\sqrt {3169}[/tex]

.

And, the magnitude of the is

[tex]|\vec C|=\sqrt {2^2+ 43^2}[/tex]

[tex]\Rightarrow |\vec C|=\sqrt {1853}[/tex]

.

(b) The difference between the two vectors is the difference between the corresponding components of the vectors. So, the required expression of is

[tex]\vec A - \vec C=(24 \hat i +33 \hat j) - (2 \hat i +43 \hat j)[/tex]

[tex]\Rightarrow \vec A - \vec C=24 \hat i +33 \hat j - 2 \hat i -43 \hat j[/tex]

[tex]\Rightarrow \vec A - \vec C=22 \hat i -10 \hat j[/tex]

(c) The expression of is

[tex]\vec A - \vec N=(24 \hat i +33 \hat j) - (55 \hat i -12 \hat j)[/tex]

[tex]\Rightarrow \vec A - \vec B=24 \hat i +33 \hat j - 55\hat i +12 \hat j[/tex]

[tex]\Rightarrow \vec A - \vec B=-31 \hat i +45 \hat j\;\cdots (i)[/tex]

The magnitude of is

[tex]|\vec A - \vec B|=\sqrt {(-31)^2+55^2}[/tex]

[tex]\Rightarrow |\vec A - \vec B|=\sqrt {3986}[/tex]

[tex]\Rightarrow |\vec A - \vec B|=63.13[/tex]

Now, if a vector [tex]\vec V= -\alpha \hat i +\beta \hat j[/tex] in 3rd quadrant having direction [tex]\theta[/tex] with respect to [tex]\hat i[/tex] direction, than

in the anti-clockwise direction.

Here, from equation (i), for the vector [tex]\vec A - \vec C[/tex], [tex]\alpha=31[/tex] and [tex]\beta=45[/tex].

[tex]\Rightarrow \theta = \pi-\tan ^{-1}\left(\frac {45}{31}\right)[/tex]

180°-55.44° [as \pi radian= 180°]

124.56° in the anti-clockwise direction.

(d) Vector diagrams for [tex]\vec A +\vec B[/tex] and [tex]\vec A - \vec B[/tex] has been shown  

in the figure(b) and figure(c) recpectively.

Vector [tex]\vec A - \vec B[/tex] is in 3rd quadrant as calculated in part (c).

While Vector [tex]\vec A +\vec B=(24 \hat i +33 \hat j)+(55 \hat i -12 \hat j)[/tex]

[tex]\Rightarrow \vec A +\vec B=79 \hat i +21 \hat j[/tex], which is in 1st quadrant as both the components are position has been shown in figure(b).

Ver imagen Ritz01