/ector questions: 1) The angle between two vectors overrightarrow (A) and bar (B) , that are starting from the same point is 60^circ If bar (A)=7i-j+5k and vert Bvert =7 Then the scalar product of these two vectors is: 150.3 b) 36.7 c) 63.6 d) 262 e) 52.5

To find the scalar product of two vectors, we can use the formula:<br /><br />$\overrightarrow{A} \cdot \overrightarrow{B} = \vert \overrightarrow{A} \vert \cdot \vert \overrightarrow{B} \vert \cdot \cos(\theta)$<br /><br />where $\theta$ is the angle between the two vectors.<br /><br />Given that the angle between $\overrightarrow{A}$ and $\overrightarrow{B}$ is $60^{\circ}$, we can substitute the values into the formula:<br /><br />$\overrightarrow{A} \cdot \overrightarrow{B} = \vert \overrightarrow{A} \vert \cdot \vert \overrightarrow{B} \vert \cdot \cos(60^{\circ})$<br /><br />We are given that $\vert \overrightarrow{B} \vert = 7$ and $\overrightarrow{A} = 7i - j + 5k$. To find the magnitude of $\overrightarrow{A}$, we can use the formula:<br /><br />$\vert \overrightarrow{A} \vert = \sqrt{A_x^2 + A_y^2 + A_z^2}$<br /><br />where $A_x$, $A_y$, and $A_z$ are the components of $\overrightarrow{A}$.<br /><br />Substituting the values, we have:<br /><br />$\vert \overrightarrow{A} \vert = \sqrt{7^2 + (-1)^2 + 5^2} = \sqrt{49 + 1 + 25} = \sqrt{75} = 5\sqrt{3}$<br /><br />Now, we can substitute the values into the scalar product formula:<br /><br />$\overrightarrow{A} \cdot \overrightarrow{B} = 5\sqrt{3} \cdot 7 \cdot \cos(60^{\circ})$<br /><br />Since $\cos(60^{\circ}) = \frac{1}{2}$, we have:<br /><br />$\overrightarrow{A} \cdot \overrightarrow{B} = 5\sqrt{3} \cdot 7 \cdot \frac{1}{2} = \frac{35\sqrt{3}}{2}$<br /><br />Therefore, the scalar product of the two vectors is $\frac{35\sqrt{3}}{2}$, which is approximately 52.5.<br /><br />So, the correct answer is e) 52.5.