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Notely Maths 12
Class 12 Maths
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7.1 Q-6

Question Statement

Find the unit vector in the direction of the given vectors:

(i) V‾=2i‾−j‾\underline{V} = 2 \underline{\mathbf{i}} - \underline{\mathbf{j}}
(ii) V‾=12i‾−32j‾\underline{V} = \frac{1}{2} \underline{i} - \frac{\sqrt{3}}{2} \underline{j}
(iii) V‾=−32i‾−12j‾\underline{V} = -\frac{\sqrt{3}}{2} \underline{i} - \frac{1}{2} \underline{j}

Background and Explanation

A unit vector is a vector with a magnitude of 1. To find a unit vector in the direction of a given vector V‾\underline{V}, we use the formula: U‾=V‾∣V‾∣\underline{U} = \frac{\underline{V}}{|\underline{V}|}

Where ∣V‾∣|\underline{V}| is the magnitude of the vector V‾\underline{V}. The magnitude of a 2D vector V‾=xi^+yj^\underline{V} = x\hat{i} + y\hat{j} is given by: ∣V‾∣=x2+y2|\underline{V}| = \sqrt{x^2 + y^2}

After finding the magnitude, we divide each component of V‾\underline{V} by its magnitude to get the unit vector U‾\underline{U}.

Solution

(i) V‾=2i‾−j‾\underline{V} = 2 \underline{\mathbf{i}} - \underline{\mathbf{j}}

  1. Find the magnitude of V‾\underline{V}:
    ∣V‾∣=(2)2+(−1)2=4+1=5|\underline{V}| = \sqrt{(2)^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5}

  2. Find the unit vector U‾\underline{U}:
    U‾=2i^−j^5\underline{U} = \frac{2\hat{i} - \hat{j}}{\sqrt{5}}

  3. Simplify the result:
    U‾=25i^−15j^\underline{U} = \frac{2}{\sqrt{5}} \hat{i} - \frac{1}{\sqrt{5}} \hat{j}

Thus, the unit vector in the direction of V‾\underline{V} is:
U‾=25i^−15j^\underline{U} = \frac{2}{\sqrt{5}} \hat{i} - \frac{1}{\sqrt{5}} \hat{j}

(ii) V‾=12i‾−32j‾\underline{V} = \frac{1}{2} \underline{i} - \frac{\sqrt{3}}{2} \underline{j}

  1. Find the magnitude of V‾\underline{V}:
    ∣V‾∣=(12)2+(32)2=14+34=1=1|\underline{V}| = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1

  2. Find the unit vector U‾\underline{U}:
    Since the magnitude of V‾\underline{V} is already 1, the unit vector is just V‾\underline{V} itself:
    U‾=12i^−32j^\underline{U} = \frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{j}

Thus, the unit vector in the direction of V‾\underline{V} is:
U‾=12i^−32j^\underline{U} = \frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{j}

(iii) V‾=−32i‾−12j‾\underline{V} = -\frac{\sqrt{3}}{2} \underline{i} - \frac{1}{2} \underline{j}

  1. Find the magnitude of V‾\underline{V}:
    ∣V‾∣=(−32)2+(−12)2=34+14=1=1|\underline{V}| = \sqrt{\left(-\frac{\sqrt{3}}{2}\right)^2 + \left(-\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4} + \frac{1}{4}} = \sqrt{1} = 1

  2. Find the unit vector U‾\underline{U}:
    Since the magnitude of V‾\underline{V} is already 1, the unit vector is simply V‾\underline{V}:
    U‾=−32i^−12j^\underline{U} = -\frac{\sqrt{3}}{2} \hat{i} - \frac{1}{2} \hat{j}

Thus, the unit vector in the direction of V‾\underline{V} is:
U‾=−32i^−12j^\underline{U} = -\frac{\sqrt{3}}{2} \hat{i} - \frac{1}{2} \hat{j}

Key Formulas or Methods Used

  • Magnitude of a vector:
    ∣V‾∣=x2+y2|\underline{V}| = \sqrt{x^2 + y^2}
    Where xx and yy are the components of the vector.

  • Unit vector:
    U‾=V‾∣V‾∣\underline{U} = \frac{\underline{V}}{|\underline{V}|}
    Where ∣V‾∣|\underline{V}| is the magnitude of the vector V‾\underline{V}.

Summary of Steps

  1. Find the magnitude of the given vector V‾\underline{V} using the formula:
    ∣V‾∣=x2+y2|\underline{V}| = \sqrt{x^2 + y^2}

  2. Divide each component of V‾\underline{V} by its magnitude to find the unit vector:
    U‾=V‾∣V‾∣\underline{U} = \frac{\underline{V}}{|\underline{V}|}

  3. Simplify the expression to obtain the unit vector in the desired direction.