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 problem asks for the value of \(|\vec{a} \times \vec{d}|^2\), given the vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) and two conditions that define the vector \(\vec{d}\).

Concept Used:

This problem utilizes several fundamental properties of vector algebra:

1. Cross Product Property: If the cross product of two non-zero vectors is the zero vector, i.e., \(\vec{u} \times \vec{v} = \vec{0}\), then the vectors \(\vec{u}\) and \(\vec{v}\) are parallel (or collinear). This means one vector can be expressed as a scalar multiple of the other, \(\vec{v} = \lambda \vec{u}\) for some scalar \(\lambda\).
2. Dot Product (Scalar Product): The dot product of two vectors \(\vec{u} = u_1\hat{i} + u_2\hat{j} + u_3\hat{k}\) and \(\vec{v} = v_1\hat{i} + v_2\hat{j} + v_3\hat{k}\) is given by \(\vec{u} \cdot \vec{v} = u_1v_1 + u_2v_2 + u_3v_3\).
3. Cross Product Calculation: The cross product can be calculated using the determinant form: \[ \vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix} \]
4. Magnitude of a Vector: The squared magnitude of a vector \(\vec{w} = w_1\hat{i} + w_2\hat{j} + w_3\hat{k}\) is \(|\vec{w}|^2 = w_1^2 + w_2^2 + w_3^2\).

Step-by-Step Solution:

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

The first condition is \(\vec{b} \times \vec{d} = \vec{c} \times \vec{d}\). We can rearrange this equation:

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

Using the distributive property of the cross product:

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

This implies that the vector \(\vec{d}\) is parallel to the vector \((\vec{b} - \vec{c})\). Therefore, \(\vec{d}\) can be written as a scalar multiple of \((\vec{b} - \vec{c})\).

\[ \vec{d} = \lambda (\vec{b} - \vec{c}) \quad \text{for some scalar } \lambda \]

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

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

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

Step 3: Use the second condition to find the scalar \(\lambda\).

We now have \(\vec{d} = \lambda (\hat{i} - 2\hat{j} + \hat{k})\). The second condition is \(\vec{a} \cdot \vec{d} = 4\). We are given \(\vec{a} = \hat{i} + 2\hat{j} + \hat{k}\).

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

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

Substitute the value of \(\lambda = -2\) back into the expression for \(\vec{d}\):

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

Step 5: Calculate 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 the final value \(|\vec{a} \times \vec{d}|^2\).

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

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

The value of \(|\vec{a} \times \vec{d}|^2\) is 128.

Answer 2

Given 
\( \vec{b} \times \vec{d} = \vec{c} \times \vec{d} \). 
\( \vec{b} \times \vec{d} - \vec{c} \times \vec{d} = \vec{0} \) 
\( (\vec{b} - \vec{c}) \times \vec{d} = \vec{0} \) 
This implies that \( \vec{d} \) is parallel to \( \vec{b} - \vec{c} \). 
\( \vec{b} - \vec{c} = (3\hat{i} - 3\hat{j} + 3\hat{k}) - (2\hat{i} - \hat{j} + 2\hat{k}) = (3 - 2)\hat{i} + (-3 - (-1))\hat{j} + (3 - 2)\hat{k} = \hat{i} - 2\hat{j} + \hat{k} \) 
So, \( \vec{d} = \lambda (\vec{b} - \vec{c}) = \lambda (\hat{i} - 2\hat{j} + \hat{k}) \) 
for some scalar \( \lambda \). Given \( \vec{a} \cdot \vec{d} = 4 \). 
\( (\hat{i} + 2\hat{j} + \hat{k}) \cdot (\lambda (\hat{i} - 2\hat{j} + \hat{k})) = 4 \) 
\( \lambda ((\hat{i} + 2\hat{j} + \hat{k}) \cdot (\hat{i} - 2\hat{j} + \hat{k})) = 4 \) 
\( \lambda (1(1) + 2(-2) + 1(1)) = 4 \) \( \lambda (1 - 4 + 1) = 4 \) \( \lambda (-2) = 4 \) \( \lambda = -2 \) 
Now we can find \( \vec{d} \): \( \vec{d} = -2 (\hat{i} - 2\hat{j} + \hat{k}) = -2\hat{i} + 4\hat{j} - 2\hat{k} \) 
We need to find \( |\vec{a} \times \vec{d}|^2 \). First, calculate \( \vec{a} \times \vec{d} \): \[ \vec{a} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} 1 & 2 & 1 -2 & 4 & -2 \end{vmatrix} = \hat{i}(2(-2) - 1(4)) - \hat{j}(1(-2) - 1(-2)) + \hat{k}(1(4) - 2(-2)) \] \[ \vec{a} \times \vec{d} = \hat{i}(-4 - 4) - \hat{j}(-2 + 2) + \hat{k}(4 + 4) = -8\hat{i} - 0\hat{j} + 8\hat{k} = -8\hat{i} + 8\hat{k} \] Now, find the magnitude squared: \[ |\vec{a} \times \vec{d}|^2 = (-8)^2 + (0)^2 + (8)^2 = 64 + 0 + 64 = 128 \] 
Alternatively, using the identity \( |\vec{a} \times \vec{d}|^2 + (\vec{a} \cdot \vec{d})^2 = |\vec{a}|^2 |\vec{d}|^2 \): 
\( |\vec{a}|^2 = 1^2 + 2^2 + 1^2 = 1 + 4 + 1 = 6 \) \( |\vec{d}|^2 = (-2)^2 + (4)^2 + (-2)^2 = 4 + 16 + 4 = 24 \) \( (\vec{a} \cdot \vec{d})^2 = (4)^2 = 16 \) \( |\vec{a} \times \vec{d}|^2 = |\vec{a}|^2 |\vec{d}|^2 - (\vec{a} \cdot \vec{d})^2 = 6 \times 24 - 16 = 144 - 16 = 128 \)