Question

for v1=2i-3 j and v2 =-6i+5j determine the magnitude and direction of v1+v2.
Ànswer ⬇⬇⬇⬇⬇⬇
$2 \sqrt{5} theta = \tan( - 1( - 1 \div 2)with \: x - axis$
​

Answer 1

Answer:

So, magnitude is $2\sqrt{5}$ and dirction of the vectors is $tan^{-1}(\frac{-1}{2})$ .

Explanation:

For v1=2i-3 j and v2 =-6i+5j determine the magnitude and direction of v1+v2.

Since given data in the question is:

$V_{1}=2\hat{i}-3\hat{j}$

$V_{2}=-6\hat{i}+5\hat{j}$

$V_{1} +V_{2} =-4\hat{i}+2\hat{j}$

So, the magnitude of $V_{1} +V_{2}$=$\sqrt{(-4)^{2}+2^{2}}$

$V_{1} +V_{2}$=$\sqrt{20}$

$V_{1} +V_{2}$=$2\sqrt{5}$

for finding the direction we know that the

$tan\alpha =\frac{y}{x}$

So,$tan\alpha =\frac{2}{-4}$

$\alpha =tan^{-1} \frac{-1}{2}$

So, magnitude is $2\sqrt{5}$ and direction of the vectors is $tan^{-1}(\frac{1}{2})$

Answer 2

Answer:

Magnitude = $2\sqrt 5$.

Direction = $153.43^\circ$ measured counterclockwise from positive x axis.

Explanation:

Given:

• $\vec v_1 = 2\hat i-3\hat j$.
• $\vec v_2 = -6\hat i+5\hat j$.

$\vec v_1 +\vec v_2 = (2\hat i-3\hat j)+(-6\hat i+5\hat j)\\=(2-6)\hat i+(-3+5)\hat j\\=-4\hat i+2\hat j.$

$\hat i\ \text{and } \hat j$ are unit vectors along the x and y axes respectively.

Thus,

Magnitude of $\vec v_1 +\vec v_2$ = $\sqrt{(-4)^2+(2)^2}=\sqrt {20} = 2\sqrt 5$

Direction of $\vec v_1 +\vec v_2$ with respect to x axis is given by

$\tan\theta = \dfrac{(\vec v_1 + \vec v_2)_y}{(\vec v_1 + \vec v_2)_x}=\dfrac{2}{-4}=\dfrac{-1}{2}\\\theta=\tan^{-1}\dfrac{-1}{2}=-26.56^\circ.$

Since, the x component of $\vec v_1 +\vec v_2$ is negative and its y component is positive therefore $\vec v_1 +\vec v_2$ lies in second quadrant.

Thus, its angle with respect to positive x axis, counterclockwise = $180^\circ-26.56^\circ = 153.43^\circ.$