To find the dot product of two vectors [tex]\(\vec{A} = 4\vec{i} + 6\vec{j} + 3\vec{k}\)[/tex] and [tex]\(\vec{B} = 3\vec{i} + 4\vec{j} + 5\vec{k}\)[/tex], we use the formula for the dot product of two vectors:
[tex]\[
\vec{A} \cdot \vec{B} = A_i B_i + A_j B_j + A_k B_k
\][/tex]
Where:
- [tex]\(A_i\)[/tex], [tex]\(A_j\)[/tex], and [tex]\(A_k\)[/tex] are the components of vector [tex]\(\vec{A}\)[/tex].
- [tex]\(B_i\)[/tex], [tex]\(B_j\)[/tex], and [tex]\(B_k\)[/tex] are the components of vector [tex]\(\vec{B}\)[/tex].
In this problem, we have:
[tex]\[
A_i = 4, \quad A_j = 6, \quad A_k = 3
\][/tex]
[tex]\[
B_i = 3, \quad B_j = 4, \quad B_k = 5
\][/tex]
Now, substitute these values into the formula:
[tex]\[
\vec{A} \cdot \vec{B} = (4 \cdot 3) + (6 \cdot 4) + (3 \cdot 5)
\][/tex]
Calculate each term individually:
[tex]\[
4 \cdot 3 = 12
\][/tex]
[tex]\[
6 \cdot 4 = 24
\][/tex]
[tex]\[
3 \cdot 5 = 15
\][/tex]
Now add these results together:
[tex]\[
\vec{A} \cdot \vec{B} = 12 + 24 + 15 = 51
\][/tex]
Therefore, the dot product [tex]\(\vec{A} \cdot \vec{B}\)[/tex] is 51. The correct answer is:
1) 51
