Question

Let $ \vec{a} = \hat{i} + 2\hat{j} + \hat{k} $, $ \vec{b} = 3\hat{i} - 3\hat{j} + 3\hat{k} $, $ \vec{c} = 2\hat{i} - \hat{j} + 2\hat{k} $ and $ \vec{d} $ be a vector such that $ \vec{b} \times \vec{d} = \vec{c} \times \vec{d} $ and $ \vec{a} \cdot \vec{d} = 4 $. Then $ |\vec{a} \times \vec{d}|^2 $ is equal to _______

Hint:

The condition \( (\vec{b} - \vec{c}) \times \vec{d} = \vec{0} \) implies that \( \vec{d} \) is parallel to \( \vec{b} - \vec{c} \), so \( \vec{d} = \lambda (\vec{b} - \vec{c}) \). Use the dot product condition to find \( \lambda \), and then calculate the cross product and its magnitude squared. Alternatively, use the vector identity relating the magnitudes of the cross product and dot product.

Answer 1

Correct Answer: 128

The objective is to determine the squared magnitude of the cross product \(|\vec{a} \times \vec{d}|^2\), given vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\), and two conditions that define \(\vec{d}\).

Key Concepts:

This problem employs the following vector algebra principles:

1. Cross Product and Parallelism: If \(\vec{u} \times \vec{v} = \vec{0}\) for non-zero vectors \(\vec{u}\) and \(\vec{v}\), then \(\vec{u}\) and \(\vec{v}\) are parallel. This means \(\vec{v} = \lambda \vec{u}\) for some scalar \(\lambda\).
2. Dot Product: For \(\vec{u} = u_1\hat{i} + u_2\hat{j} + u_3\hat{k}\) and \(\vec{v} = v_1\hat{i} + v_2\hat{j} + v_3\hat{k}\), the dot product is \(\vec{u} \cdot \vec{v} = u_1v_1 + u_2v_2 + u_3v_3\).
3. Cross Product Calculation: The cross product is computed via the determinant: \[\vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix}\]
4. Vector Magnitude: The squared magnitude of \(\vec{w} = w_1\hat{i} + w_2\hat{j} + w_3\hat{k}\) is \(|\vec{w}|^2 = w_1^2 + w_2^2 + w_3^2\).

Solution Process:

Step 1: Simplify the first condition for \(\vec{d}\).

The condition \(\vec{b} \times \vec{d} = \vec{c} \times \vec{d}\) can be rewritten as:

\[\vec{b} \times \vec{d} - \vec{c} \times \vec{d} = \vec{0}\]

Applying the distributive property yields:

\[(\vec{b} - \vec{c}) \times \vec{d} = \vec{0}\]

This indicates that \(\vec{d}\) is parallel to \((\vec{b} - \vec{c})\), thus \(\vec{d} = \lambda (\vec{b} - \vec{c})\) for a scalar \(\lambda\).

Step 2: Compute the vector \((\vec{b} - \vec{c})\).

Given \(\vec{b} = 3\hat{i} - 3\hat{j} + 3\hat{k}\) and \(\vec{c} = 2\hat{i} - \hat{j} + 2\hat{k}\):

\[\vec{b} - \vec{c} = (3 - 2)\hat{i} + (-3 - (-1))\hat{j} + (3 - 2)\hat{k}\]\[\vec{b} - \vec{c} = \hat{i} - 2\hat{j} + \hat{k}\]

Step 3: Determine \(\lambda\) using the second condition.

With \(\vec{d} = \lambda (\hat{i} - 2\hat{j} + \hat{k})\) and \(\vec{a} = \hat{i} + 2\hat{j} + \hat{k}\), the condition \(\vec{a} \cdot \vec{d} = 4\) becomes:

\[(\hat{i} + 2\hat{j} + \hat{k}) \cdot (\lambda\hat{i} - 2\lambda\hat{j} + \lambda\hat{k}) = 4\]\[(1)(\lambda) + (2)(-2\lambda) + (1)(\lambda) = 4\]\[\lambda - 4\lambda + \lambda = 4\]\[-2\lambda = 4 \implies \lambda = -2\]

Step 4: Find the vector \(\vec{d}\).

Substituting \(\lambda = -2\):

\[\vec{d} = -2(\hat{i} - 2\hat{j} + \hat{k}) = -2\hat{i} + 4\hat{j} - 2\hat{k}\]

Step 5: Compute the cross product \(\vec{a} \times \vec{d}\).

\[\vec{a} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ -2 & 4 & -2 \end{vmatrix}\]\[= \hat{i}((2)(-2) - (1)(4)) - \hat{j}((1)(-2) - (1)(-2)) + \hat{k}((1)(4) - (2)(-2))\]\[= \hat{i}(-4 - 4) - \hat{j}(-2 + 2) + \hat{k}(4 + 4)\]\[= -8\hat{i} - 0\hat{j} + 8\hat{k} = -8\hat{i} + 8\hat{k}\]

Step 6: Calculate \(|\vec{a} \times \vec{d}|^2\).

The squared magnitude of \(-8\hat{i} + 8\hat{k}\) is:

\[|\vec{a} \times \vec{d}|^2 = (-8)^2 + (0)^2 + (8)^2\]\[= 64 + 0 + 64 = 128\]The result for \(|\vec{a} \times \vec{d}|^2\) is 128.