Multiple Choice Questions 1. 2081 Set ENW Q.No. 5 Which one of the following is angle between two vectors \( \vec{i}+\vec{j} \) and \( \vec{j}+\vec{k} \) ? \( \begin{array}{ll}\text { a. } 0^{\circ} & \text { b. } 60^{\circ} \\ \text { c. } 90^{\circ} & \text { d. } 180^{\circ}\end{array} \)

To find the angle between the two vectors \( \vec{A} = \vec{i} + \vec{j} \) and \( \vec{B} = \vec{j} + \vec{k} \), we can use the formula for the cosine of the angle \( \theta \) between two vectors: \[ \cos(\theta) = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|} \] ### Step 1: Calculate the dot product \( \vec{A} \cdot \vec{B} \) The dot product of two vectors \( \vec{A} = a_1 \vec{i} + a_2 \vec{j} + a_3 \vec{k} \) and \( \vec{B} = b_1 \vec{i} + b_2 \vec{j} + b_3 \vec{k} \) is given by: \[ \vec{A} \cdot \vec{B} = a_1 b_1 + a_2 b_2 + a_3 b_3 \] For our vectors: - \( \vec{A} = 1\vec{i} + 1\vec{j} + 0\vec{k} \) (i.e., \( a_1 = 1, a_2 = 1, a_3 = 0 \)) - \( \vec{B} = 0\vec{i} + 1\vec{j} + 1\vec{k} \) (i.e., \( b_1 = 0, b_2 = 1, b_3 = 1 \)) Calculating the dot product: \[ \vec{A} \cdot \vec{B} = (1)(0) + (1)(1) + (0)(1) = 0 + 1 + 0 = 1 \] ### Step 2: Calculate the magnitudes \( |\vec{A}| \) and \( |\vec{B}| \) The magnitude of a vector \( \vec{A} = a_1 \vec{i} + a_2 \vec{j} + a_3 \vec{k} \) is given by: \[ |\vec{A}| = \sqrt{a_1^2 + a_2^2 + a_3^2} \] Calculating the magnitudes: \[ |\vec{A}| = \sqrt{1^2 + 1^2 + 0^2} = \sqrt{1 + 1 + 0} = \sqrt{2} \] \[ |\vec{B}| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{0 + 1 + 1} = \sqrt{2} \] ### Step 3: Substitute into the cosine formula Now substituting into the cosine formula: \[ \cos(\theta) = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2} \] ### Step 4: Find the angle \( \theta \) To find \( \theta \), we take the inverse cosine: \[ \theta = \cos^{-1}\left(\frac{1}{2}\right) \] This gives: \[ \theta = 60^{\circ} \] ### Conclusion The angle between the vectors \( \vec{i} + \vec{j} \) and \( \vec{j} + \vec{k} \) is \( 60^{\circ} \). Thus, the correct answer is **b. \( 60^{\circ} \)**.