Question

In a meeting of mimes, mime 1 goes through a displacement d1 = (6.00 m)i + (5.74 m)j and and mime 2 goes through a displacement d2 = (-2.92 m)i + (2.9 m)j. What are
a. |d1 × d2|,
b. d1 · d2,
c. (d1 + d2) · d2,
d. The component of d1 along the direction of d2?

Answer 1

a) Vector product: |d1 × d2| = (34.2 m) k

b) Scalar product: d1 · d2 = -0.874 m

c) (d1 + d2) · d2 = 16.1 m

d) Component of d1 along direction of d2: -0.21 m

Explanation:

a)

In this part, we want to calculate

|d1 × d2|

Which is the vector product between the two displacements d1 and d2.

The two vectors are:

d1 = (6.00 m)i + (5.74 m)j

d2 = (-2.92 m)i + (2.9 m)j

The vector product of two vectors $(a_x,a_y,a_z)$ and $(b_x,b_y,b_z)$ is also a vector which has components:

$r=(r_x,r_y,r_z)\\r_x=a_yb_z-a_zb_y\\r_y=a_zb_x-a_xb_z\\r_z=a_xb_y-a_yb_x$

We notice immediatly that in this problem ,the two vectors d1 and d2 lie in the x-y plane, so they do not have components in zero. Therefore, the vector product has only one component, which is the one in z, and it is:

$r_z=(6.00)(2.9)-(5.74)(-2.92))=34.2 m$

Therefore, the vector product of d1 and d2 is:

|d1 × d2| = (34.2 m) k

b)

In this case, we want to calculate

d1 · d2

Which is the scalar product between the two displacements.

The scalar product of two vectors $(a_x,a_y,a_z)$ and $(b_x,b_y,b_z)$ is a scalar given by:

$a \cdot b = a_x b_x + a_y b_y + a_z b_z$

In this problem,

d1 = (6.00 m)i + (5.74 m)j

d2 = (-2.92 m)i + (2.9 m)j

Therefore, the scalar product between the two vectors is:

$d_1 \cdot d_2 = (6.00)(-2.92)+(5.74)(2.9)=-0.87m$

c)

In this  case, we want to calculate

(d1 + d2) · d2

Which means that first we have to calculate the resultant displacement d1 + d2, and then calculate the scalar product of the resultant vector with d2.

Given two vectors  $(a_x,a_y,a_z)$ and $(b_x,b_y,b_z)$, the resultant vector is also a vector given by

$r=(r_x,r_y,r_z)\\r_x=a_x+b_x\\r_y=a_y+b_y\\r_z=a_z+b_z$

In this case,

d1 = (6.00 m)i + (5.74 m)j

d2 = (-2.92 m)i + (2.9 m)j

So the resultant vector is

$r_x=6.00+(-2.92)=3.08 m\\r_y = 5.74+2.9=8.64 m$

So

$(d_1+d_2)=(3.08 m,8.64 m)$

And calculating the scalar product with d2, we find:

$(d_1 + d_2)\cdot d_2 = (3.08)(-2.92)+(8.64)(2.9)=16.1 m$

d)

The component of a vector a along another vector b is given by

$a_b = \frac{a\cdot b}{|b|}$

where$a\cdot b$ is the scalar product between and b

$|b|$ is the magnitude of vector b

In this problem, we have the two vectors

d1 = (6.00 m)i + (5.74 m)j

d2 = (-2.92 m)i + (2.9 m)j

We want to find the component of d1 along the direction of d2.

We already calculated the scalar product of the two vectors in part b):

d1 · d2 = -0.874 m

The magnitude of a vector b is given by

$|b|=\sqrt{b_x^2+b_y^2+b_z^2}$

So, for vector d2,

$|d_2|=\sqrt{(-2.92)^2+(2.9)^2}=4.1 m$

Now we can calculate the component of d1 along d2:

$d_1_{d_2}=\frac{d_1 \cdot d_2}{|d_2|}=\frac{-0.874}{4.1}=-0.21 m$

Learn more about operations with vectors:

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