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 [tex](a_x,a_y,a_z)[/tex] and [tex](b_x,b_y,b_z)[/tex] is also a vector which has components:
[tex]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[/tex]
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:
[tex]r_z=(6.00)(2.9)-(5.74)(-2.92))=34.2 m[/tex]
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 [tex](a_x,a_y,a_z)[/tex] and [tex](b_x,b_y,b_z)[/tex] is a scalar given by:
[tex]a \cdot b = a_x b_x + a_y b_y + a_z b_z[/tex]
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:
[tex]d_1 \cdot d_2 = (6.00)(-2.92)+(5.74)(2.9)=-0.87m[/tex]
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 [tex](a_x,a_y,a_z)[/tex] and [tex](b_x,b_y,b_z)[/tex], the resultant vector is also a vector given by
[tex]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[/tex]
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
[tex]r_x=6.00+(-2.92)=3.08 m\\r_y = 5.74+2.9=8.64 m[/tex]
So
[tex](d_1+d_2)=(3.08 m,8.64 m)[/tex]
And calculating the scalar product with d2, we find:
[tex](d_1 + d_2)\cdot d_2 = (3.08)(-2.92)+(8.64)(2.9)=16.1 m[/tex]
d)
The component of a vector a along another vector b is given by
[tex]a_b = \frac{a\cdot b}{|b|}[/tex]
where[tex]a\cdot b[/tex] is the scalar product between and b
[tex]|b|[/tex] 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
[tex]|b|=\sqrt{b_x^2+b_y^2+b_z^2}[/tex]
So, for vector d2,
[tex]|d_2|=\sqrt{(-2.92)^2+(2.9)^2}=4.1 m[/tex]
Now we can calculate the component of d1 along d2:
[tex]d_1_{d_2}=\frac{d_1 \cdot d_2}{|d_2|}=\frac{-0.874}{4.1}=-0.21 m[/tex]
Learn more about operations with vectors:
brainly.com/question/2927458
brainly.com/question/2088577
brainly.com/question/1592430
#LearnwithBrainly
