Question 16 of 18intermediate🔧 ApplyNumerical3 marks

i and j are unit vectors along x- and y-axis respectively. What is the magnitude and direction of the vectors i + j, and i - j? What are the components of a vector A = 2i + 3j along the directions of i + j and i - j? [You may use graphical method]

Correct Answer

(component_along_i-j) -1/2\sqrt{2}

(component_along_i+j) 5/2\sqrt{2}

(magnitude_direction_i-j) Magnitude = 2\sqrt{2}, Direction = -45° (or 315°) with x-axis

(magnitude_direction_i+j) Magnitude = 2\sqrt{2}, Direction = 45° with x-axis

Exercise: EXERCISES | Q: 3.19 | (Page 22)
For More Understanding

Explanation

This question tests understanding of vector magnitude, direction, and resolving vectors along non-standard directions. The context provides the formula A = Ax2+Ay2\sqrt{Ax² + Ay²} for magnitude and tan θ = Ay/Ax for direction. For finding components along arbitrary directions, we use the dot product method with unit vectors.

Solution Steps

  1. Step 1: Find magnitude and direction of (î + ĵ)

  2. Given: Ax = 1, Ay = 1

  3. Using A = Ax2+Ay2\sqrt{Ax² + Ay²} from Eq. (3.14): |î + ĵ| = 12+12\sqrt{1² + 1²} = 2\sqrt{2}

  4. Using tan θ = Ay/Ax from Eq. (3.15): tan θ = 1/1 = 1, so θ = 45°

  6. Step 2: Find magnitude and direction of (î - ĵ)

  7. Given: Ax = 1, Ay = -1

  8. Magnitude: |î - ĵ| = 12+(−1\sqrt{1² + (-1}²) = 2\sqrt{2}

  9. Direction: tan θ = -1/1 = -1, so θ = -45° (or 315°) with x-axis

  11. Step 3: Find unit vectors along (î + ĵ) and (î - ĵ)

  12. Unit vector along (î + ĵ): n₁ = (î + ĵ)/2\sqrt{2}

  13. Unit vector along (î - ĵ): n₂ = (î - ĵ)/2\sqrt{2}

  15. Step 4: Find components of A = 2î + 3ĵ along these directions

  16. Component along (î + ĵ) = A · n₁ = (2î + 3ĵ) · (î + ĵ)/2\sqrt{2} = (2 + 3)/2\sqrt{2} = 5/2\sqrt{2}

  17. Component along (î - ĵ) = A · n₂ = (2î + 3ĵ) · (î - ĵ)/2\sqrt{2} = (2 - 3)/2\sqrt{2} = -1/2\sqrt{2}

PreviousNext