Question

Let\( |\vec{a}|=2, |\vec{b}|=3\) and the angle between the vectors \(\vec{a}\) and \(\vec{b}\) be \(\frac{π}{4}\). Then \(|(\vec{a}+2\vec{b})×(2\vec{a}-3\vec{b})|^2\) is equal to

• A. 441
• B. 882
• C. 482
• D. 841

Answer 1

The Correct Option is B

We are given that the magnitudes of the vectors \( \mathbf{a} \) and \( \mathbf{b} \) are \( |\mathbf{a}| = 2 \) and \( |\mathbf{b}| = 3 \), and the angle between them is \( \frac{\pi}{4} \). We are required to find: \[ \left| (\mathbf{a} + 2\mathbf{b}) \times (2\mathbf{a} - 3\mathbf{b}) \right|^2 \] Step 1: Using the dot product formula for the cosine of the angle between the vectors. From the given information, we know: \[ \cos \left( \frac{\pi}{4} \right) = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|} \] Since \( \cos \left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}} \), we can calculate the dot product: \[ \frac{1}{\sqrt{2}} = \frac{\mathbf{a} \cdot \mathbf{b}}{2 \times 3} \quad \Rightarrow \quad \mathbf{a} \cdot \mathbf{b} = 3\sqrt{2} \] Step 2: Compute the cross product. Let \( \mathbf{p} = \mathbf{a} + 2\mathbf{b} \) and \( \mathbf{q} = 2\mathbf{a} - 3\mathbf{b} \). The squared magnitude of \( \mathbf{p} \) is: \[ |\mathbf{p}|^2 = |\mathbf{a}|^2 + 4|\mathbf{b}|^2 + 4(\mathbf{a} \cdot \mathbf{b}) \] \[ = 4 + 36 + 12\sqrt{2} = 40 + 12\sqrt{2} \] The squared magnitude of \( \mathbf{q} \) is: \[ |\mathbf{q}|^2 = 4|\mathbf{a}|^2 + 9|\mathbf{b}|^2 - 12(\mathbf{a} \cdot \mathbf{b}) \] \[ = 16 + 81 - 36\sqrt{2} = 97 - 36\sqrt{2} \] Now compute the dot product \( \mathbf{p} \cdot \mathbf{q} \): \[ \mathbf{p} \cdot \mathbf{q} = 2|\mathbf{a}|^2 - 6|\mathbf{b}|^2 + \mathbf{a} \cdot \mathbf{b} \] \[ = 8 - 54 + 3\sqrt{2} = -46 + 3\sqrt{2} \] Step 3: Calculate the magnitude of the cross product. We now use the formula for the magnitude of the cross product: \[ |\mathbf{p} \times \mathbf{q}| = \sqrt{|\mathbf{p}|^2 |\mathbf{q}|^2 - (\mathbf{p} \cdot \mathbf{q})^2} \] Substitute the values: \[ |\mathbf{p} \times \mathbf{q}| = \sqrt{(40 + 12\sqrt{2})(97 - 36\sqrt{2}) - (-46 + 3\sqrt{2})^2} \] First, expand the product \( (40 + 12\sqrt{2})(97 - 36\sqrt{2}) \): \[ = 40 \times 97 + 40 \times (-36\sqrt{2}) + 12\sqrt{2} \times 97 + 12\sqrt{2} \times (-36\sqrt{2}) \] \[ = 3880 - 1440\sqrt{2} + 1164\sqrt{2} - 864 = 3016 - 276\sqrt{2} \] Now expand \( (-46 + 3\sqrt{2})^2 \): \[ = (-46)^2 + 2(-46)(3\sqrt{2}) + (3\sqrt{2})^2 = 2116 - 276\sqrt{2} + 18 = 2134 - 276\sqrt{2} \] Now subtract the two expressions: \[ 3016 - 276\sqrt{2} - (2134 - 276\sqrt{2}) = 3016 - 276\sqrt{2} - 2134 + 276\sqrt{2} = 882 \] Thus, the final result is: \[ |\mathbf{p} \times \mathbf{q}|^2 = 882 \]