Question

Vector A with arrow has a magnitude of 64 units and points due west, while vector B with arrow has the same magnitude and points due south. Specify the direction relative to due west.
Find the magnitude and direction of A with arrow and B with arrow.

Answer 1

Answer:

The magnitude is 90.5.

The direction of  $(\vec{A}+ \vec{B})$ is 45° south of west.

The direction of  $(\vec{A} - \vec{B})$ is 45° north of west.

Step-by-step explanation:

Let suppose the direction of unit vector as follows:

$\hat{i}$ in the eastern direction

$\hat{j}$ in the northern direction

Given:

Vector A with arrow has a magnitude of 64 units and points due west, while vector B with arrow has the same magnitude and points due south.

Therefore,

$\vec{A} = -64 \hat{i}$

$|\vec{A}| = 64$

And

$\vec{B} = -64 \hat{j}$

$|\vec{B}| = 64$

Formula used:

$The\ magnitude\ of\ vector\ (\vec{A}+ \vec{B}) = \sqrt{|A|^{2}+|B|^{2}}$

And the direction is given by:

$tan\theta = \frac{|A|}{|B|}$

Now,

The magnitude of $(\vec{A}+ \vec{B})$:

$= \sqrt{|A|^{2}+|B|^{2}}$

$= \sqrt{64^{2}+64^{2}}$

$= 64\sqrt{2}$

$= 90.5$

∴ The magnitude is 90.5.

And,

The direction relative to west is

$tan\theta = \frac{|A|}{|B|}$

θ = tan⁻¹ (B/A) = tan⁻¹ (64/64) = tan⁻¹ 1 = 45°  south of west

∴ The direction is 45° south of west.

Again,

The magnitude of $(\vec{A} - \vec{B})$:

$= \sqrt{|A|^{2}+|B|^{2}}$

$= \sqrt{64^{2}+64^{2}}$

$= 64\sqrt{2}$

$= 90.5$

∴ The magnitude is 90.5.

And,

The direction relative to west is

$tan\theta = \frac{|A|}{|B|}$

θ = tan⁻¹ (B/A) = tan⁻¹ (64/64) = tan⁻¹ 1 = 45° north of west.

∴ The direction is 45° north of west.