Question

A and B are two vectors such that A is 2 m long and is 60° above the x-axis in the first quadrant.
Bis 2 m long and is 60° below the x-axis in the fourth quadrant.
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(a) Draw diagram to show A + B, A- B, and B - A
(b) What is the magnitude and direction of A + B, A- B, and B - A.​

Answer 1

Given :-

◉ A is a vector which is of magnitude 2 m and is above the x-axis at an angle of 60° in the first quadrant.

◉ B is a vector which is also of magnitude 2 m and is below the x-axis at an angle of 60° in the fourth quadrant.

To Find or Show :-

◉ Diagram of A + B, A - B and B - A

◉ Magnitude and direction of A + B, A - B and B - A

Solution :-

[i]

For the diagrams, Kindly refer to the attachment.

[ii]

We have,

• |A| = 2 and at an angle of 60° above x-axis in first quadrant.
• |B| = 2 and at an angle of 60° below the x-axis in fourth quadrant.

Since, First quadrant is just above the fourth quadrant.

∴ Angle between A & B = 60° + 60° ⇒ 120°

Now, Let us find A + B,

⇒ |A + B| = √(A² + B² + 2ABcosθ)

⇒ |A + B| = √(2² + 2² + 2×2×2×-1/2)

⇒ |A + B| = √(4 + 4 - 4)

⇒ |A + B| = √4

⇒ |A + B| = 2

Hence, The magnitude of A + B is also 2.

Direction of A + B would be along the positive x-axis because the y-component would be cancelled out.

Now, We have to find the magnitude and direction of A - B

⇒ |A - B| = √(A² + B² - 2ABcosθ)

⇒ |A - B| = √(2² + 2² - 2×2×2cos×60°

⇒ |A - B| = √(4 + 4 - 4)

⇒ |A - B| = √4

⇒ |A - B| = 2

Therefore, The magnitude of A - B is also 2.

Direction of A - B would be along the positive y-axis because the x-component would be cancelled out.

Finally, We have to find the magnitude and direction of B - A.

⇒ |B - A| = √(B² + A² - 2ABcosθ)

⇒ |B - A| = √(2² + 2² - 2×2×2×cos60°)

⇒ |B - A| = √(4 + 4 - 4)

⇒ |B - A| = √4

⇒ |B - A| = 2

We again got the same magnitude i.e., 2 and the direction of B - A would be along the negative y-axis because the x-component would be cancelled out.