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 vectors \( \vec{i}+\vec{j} \) and \( \vec{j}+\vec{k} \), we first need to calculate the dot product of the two vectors. The dot product is given by: \[ (\vec{i}+\vec{j}) \cdot (\vec{j}+\vec{k}) = \vec{i} \cdot \vec{j} + \vec{i} \cdot \vec{k} + \vec{j} \cdot \vec{j} + \vec{j} \cdot \vec{k} \] Since \( \vec{i} \cdot \vec{j} = 0 \), \( \vec{i} \cdot \vec{k} = 0 \), \( \vec{j} \cdot \vec{j} = 1 \), and \( \vec{j} \cdot \vec{k} = 0\), we get: \[ 0 + 0 + 1 + 0 = 1 \] Next, we compute the magnitudes of the vectors: \[ |\vec{i}+\vec{j}| = \sqrt{1^2 + 1^2} = \sqrt{2} \] \[ |\vec{j}+\vec{k}| = \sqrt{1^2 + 1^2} = \sqrt{2} \] Now we can use the dot product formula relating to the angle \( \theta \): \[ (\vec{i}+\vec{j}) \cdot (\vec{j}+\vec{k}) = |\vec{i}+\vec{j}| \cdot |\vec{j}+\vec{k}| \cdot \cos \theta \] Substituting in the values we found: \[ 1 = \sqrt{2} \cdot \sqrt{2} \cdot \cos \theta \] This simplifies to: \[ 1 = 2 \cos \theta \implies \cos \theta = \frac{1}{2} \] Thus, \[ \theta = 60^{\circ} \] So the answer is **b. 60°**.