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Find the angle between the vectors a+b a...

Find the angle between the vectors a+b and a-b, if `a=2hat(i)-hat(j)+3hat(k) and b=3hat(i)+hat(j)-2hat(k)`.

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To find the angle between the vectors \( \mathbf{a} + \mathbf{b} \) and \( \mathbf{a} - \mathbf{b} \), we will follow these steps: ### Step 1: Calculate \( \mathbf{a} + \mathbf{b} \) Given: \[ \mathbf{a} = 2\hat{i} - \hat{j} + 3\hat{k} \] \[ \mathbf{b} = 3\hat{i} + \hat{j} - 2\hat{k} \] Now, we calculate \( \mathbf{a} + \mathbf{b} \): \[ \mathbf{a} + \mathbf{b} = (2\hat{i} + 3\hat{i}) + (-\hat{j} + \hat{j}) + (3\hat{k} - 2\hat{k}) \] \[ = (2 + 3)\hat{i} + (-1 + 1)\hat{j} + (3 - 2)\hat{k} \] \[ = 5\hat{i} + 0\hat{j} + 1\hat{k} \] \[ = 5\hat{i} + \hat{k} \] ### Step 2: Calculate \( \mathbf{a} - \mathbf{b} \) Next, we calculate \( \mathbf{a} - \mathbf{b} \): \[ \mathbf{a} - \mathbf{b} = (2\hat{i} - 3\hat{i}) + (-\hat{j} - \hat{j}) + (3\hat{k} + 2\hat{k}) \] \[ = (2 - 3)\hat{i} + (-1 - 1)\hat{j} + (3 + 2)\hat{k} \] \[ = -1\hat{i} - 2\hat{j} + 5\hat{k} \] ### Step 3: Find the dot product \( (\mathbf{a} + \mathbf{b}) \cdot (\mathbf{a} - \mathbf{b}) \) Let \( \mathbf{c} = \mathbf{a} + \mathbf{b} = 5\hat{i} + \hat{k} \) and \( \mathbf{d} = \mathbf{a} - \mathbf{b} = -\hat{i} - 2\hat{j} + 5\hat{k} \). Now we calculate the dot product: \[ \mathbf{c} \cdot \mathbf{d} = (5\hat{i} + \hat{k}) \cdot (-\hat{i} - 2\hat{j} + 5\hat{k}) \] \[ = 5 \cdot (-1) + 0 \cdot (-2) + 1 \cdot 5 \] \[ = -5 + 0 + 5 \] \[ = 0 \] ### Step 4: Use the dot product to find the angle The formula for the dot product is: \[ \mathbf{c} \cdot \mathbf{d} = |\mathbf{c}| |\mathbf{d}| \cos \theta \] Since \( \mathbf{c} \cdot \mathbf{d} = 0 \), we have: \[ |\mathbf{c}| |\mathbf{d}| \cos \theta = 0 \] This implies: \[ \cos \theta = 0 \] ### Step 5: Determine the angle \( \theta \) The angle \( \theta \) for which \( \cos \theta = 0 \) is: \[ \theta = \frac{\pi}{2} \text{ radians} \quad \text{or} \quad 90^\circ \] ### Final Answer: The angle between the vectors \( \mathbf{a} + \mathbf{b} \) and \( \mathbf{a} - \mathbf{b} \) is \( 90^\circ \). ---
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