Find the scalar and vector products of two vectors. A = (3i- 4j + 5k) and
B = ( 2i + j- 3k)
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Answers
Answer:
We have been given two vectors:
a=3\hat i-4\hat j+5\hat ka=3
i
^
−4
j
^
+5
k
^
b=-2\hat i+\hat j-3\hat kb=−2
i
^
+
j
^
−3
k
^
To find the scalar product of two vectors we will use:
a.b=a_xb_x+a_yb_y+a_zc_za.b=a
x
b
x
+a
y
b
y
+a
z
c
z
Where, i, j, and k are the components of unit vectors along 'x', 'y', and 'z'.
a_x=3,b_x=-2, a_y=-4,b_y=1, a_z=5,b_z=-3a
x
=3,b
x
=−2,a
y
=−4,b
y
=1,a
z
=5,b
z
=−3
so the scalar product is given by:
3 \times(-2) + (-4)\times1 + 5\times(-3) = -6 - 4 - 15 = -253×(−2)+(−4)×1+5×(−3)=−6−4−15=−25
So the scalar product of vectors a and b is -25−25 .
Vector product of the vectors 'a' and 'b' is given by:
(a_x, b_x, a_z)\times(a_y,b_y,b_z)=(a_y.b_z-a_z.b_y)\hat i+(a_xb_z-b_xa_z)\hat j+(a_xb_y-a_yb_x) \hat k(a
x
,b
x
,a
z
)×(a
y
,b
y
,b
z
)=(a
y
.b
z
−a
z
.b
y
)
i
^
+(a
x
b
z
−b
x
a
z
)
j
^
+(a
x
b
y
−a
y
b
x
)
k
^
Substituting the values of each component we get:
\begin{pmatrix}3&-4&5\end{pmatrix}\times \begin{pmatrix}-2&1&-3\end{pmatrix}=(7 -1 -5)(
3
−4
5
)×(
−2
1
−3
)=(7−1−5)
The vector product of 'a' and 'b' is represented by :
7\hat i-\hat j-5\hat k7
i
^
−
j
^
−5
k
^
Answer:
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