The magnitude of the vector \((\vec A+\vec B)\) is \(\boxed{16 \text{ units}}\) and the direction of the vector \((\vec A+\vec B)\) is \(\boxed{26.56^\circ\text{ anticlockwise}}\)
Further Explanation:
\(\Vec A\) points in the negative Y-direction and has a magnitude of \(7\text{ units}\), so we will show it as a line of length \(7\text{ units}\) in negative Y-direction.
\(\Vec B\) has twice the magnitude and points in the positive X-direction, so we will show it by a line of double the length as before in positive X-direction, as shown in the figure below.
According to Cartesian coordinate system, the resultant will start either from tail of \(\vec A\) and ends at head of \(\vec B\) and vice-versa.
Now, resultant will be given by
\(\begin{aligned}\vec R&=\vec A+\vec B\\\vec R&=({-7\hat j+14\hat i})\\\end{aligned}\)
Hence, the magnitude of resultant or of vector \(\vec A + \vec B\) is given by:
\(\begin{aligned}|{\vec R}|&=\sqrt{({7^2}+{14^2})}\\&=\sqrt{(49+196)}\\&=\sqrt{245}\\&=15.65{\text{ units}}\\\end{aligned}\)
It can be expressed in whole number as:
\(\boxed{|\vec R|=16\text{ units}}\)
Now, the direction of the resultant vector \(\vec R\) or of the vector \(\vec A + \vec B\) is given by:
\(\tan\theta=\dfrac{{|{\vec A}|}}{{|{\vec B}|}}\)
Now, substituting the values of vector \(\vec A\) and \(\vec B\):
\(\begin{aligned}\tan\theta&=\dfrac{7}{{14}}\\\tan\theta&=0.5\\\theta&={\tan^{-1}}(0.5)\\&=\boxed{26.56^\circ}\\\end{aligned}\)
Thus, the magnitude of the vector \((\vec A+\vec B)\) is \(\boxed{16 \text{ units}}\) and the direction of the vector \((\vec A+\vec B)\) is \(\boxed{26.56^\circ\text{ anticlockwise}}\)
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Answer Details:
Grade: High School
Subject: Physics
Chapter: Vectors and Scalars
Keywords:
Vector, scalars, vector A, Vector B, addition, resultant, unit, magnitude, direction, positive, X-axis, Y- axis, negative, 7 units, 14 units, twice the magnitude.
