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

Step 1: Compute the cross product \(\vec{d}\).

\[ \vec{d} = \vec{a} \times \vec{b} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & 1 & 1 \\ 2 & 2 & 1 \end{vmatrix} = \vec{i}(1-2) - \vec{j}(1-2) + \vec{k}(1-4) = -\vec{i} + \vec{j} - 3\vec{k} \]

Step 2: Calculate the required quantities.

Given conditions lead to a system of equations involving \(\vec{c}\), solve these using algebraic methods to find \(\vec{c}\).

Step 3: Evaluate the expression.

\[ 10 - 3\vec{b} \cdot \vec{c} + |\vec{d}| = 10 - 3(2x + 2y + z) + \sqrt{1 + 1 + 9} = 10 - 6x - 6y - 3z + \sqrt{11} \]

Compute \(\left|10 - 3\vec{b} \cdot \vec{c} + |\vec{d}|\right|^2\).