Question

Let \(\vec{a} = i + j + k\), \(\vec{b} = 2i + 2j + k\) and \(\vec{d} = \vec{a} \times \vec{b}\). If \(\vec{c}\) is a vector such that \(\vec{a} \cdot \vec{c} = |\vec{c}|\), \(\|\vec{c} - 2\vec{d}\| = 8\) and the angle between \(\vec{d}\) and \(\vec{c}\) is \(\frac{\pi}{4}\), then \(\left|10 - 3\vec{b} \cdot \vec{c} + |\vec{d}|\right|^2\) is equal to:

Hint:

For vector calculations involving cross products, determinants provide a quick method to obtain results, and verifying each step for computation errors is crucial.

Answer 1

Correct Answer: 64

Step 1: Finding \( \vec{d} = \vec{a} \times \vec{b} \)

\[ \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & 2 & 1 \end{vmatrix} \] \[ \vec{d} = \hat{i}(1 \times 1 - 1 \times 2) - \hat{j}(1 \times 1 - 1 \times 2) + \hat{k}(1 \times 2 - 1 \times 2) \] \[ \vec{d} = \hat{i}(1 - 2) - \hat{j}(1 - 2) + \hat{k}(2 - 2) \] \[ \vec{d} = -\hat{i} + \hat{j} \]

Step 2: Finding \( |\vec{d}| \)

\[ |\vec{d}| = \sqrt{(-1)^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2} \]

Step 3: Using the condition \( \vec{a} \cdot \vec{c} = |\vec{c}| \)

Let \( \vec{c} = \alpha \vec{a} + \beta \vec{d} \). \[ \vec{a} \cdot \vec{c} = \alpha (\vec{a} \cdot \vec{a}) + \beta (\vec{a} \cdot \vec{d}) \] Since \( \vec{a} \perp \vec{d} \), \( \vec{a} \cdot \vec{d} = 0 \). \[ \vec{a} \cdot \vec{c} = \alpha (\vec{a} \cdot \vec{a}) = \alpha (1^2 + 1^2 + 1^2) = 3\alpha \] Also, \( \vec{a} \cdot \vec{c} = |\vec{c}| \). \[ 3\alpha = |\vec{c}| \]

Step 4: Using the condition \( \|\vec{c} - 2\vec{d}\| = 8 \)

\[ |\vec{c} - 2\vec{d}| = 8 \] \[ |\alpha \vec{a} + \beta \vec{d} - 2\vec{d}| = 8 \] \[ \vec{c} - 2\vec{d} = \alpha \vec{a} + (\beta - 2)\vec{d} \] \[ \|\vec{c} - 2\vec{d}\| = \sqrt{(3\alpha)^2 + (\beta - 2)^2 \times 2} \] \[ 9\alpha^2 + 2(\beta - 2)^2 = 64 \]

Step 5: Using the condition about angle between \( \vec{d} \) and \( \vec{c} \)

\[ \cos \frac{\pi}{4} = \frac{\vec{d} \cdot \vec{c}}{|\vec{d}||\vec{c}|} = \frac{\beta |\vec{d}|^2}{|\vec{d}||\vec{c}|} \] \[ \frac{\beta \times 2}{\sqrt{2} \times 3\alpha} = \frac{1}{\sqrt{2}} \] \[ \frac{2\beta}{3\alpha} = 1 \implies \beta = \frac{3\alpha}{2} \]

Step 6: Substituting values and simplifying

\[ 9\alpha^2 + 2\left( \frac{3\alpha}{2} - 2 \right)^2 = 64 \] Solving this equation will give us the final result.