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

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 \( |\vec{a} \times \vec{d}|^2 + (\vec{a} \cdot \vec{d})^2 = |\vec{a}|^2 |\vec{d}|^2 \): 
\( |\vec{a}|^2 = 1^2 + 2^2 + 1^2 = 1 + 4 + 1 = 6 \) \( |\vec{d}|^2 = (-2)^2 + (4)^2 + (-2)^2 = 4 + 16 + 4 = 24 \) \( (\vec{a} \cdot \vec{d})^2 = (4)^2 = 16 \) \( |\vec{a} \times \vec{d}|^2 = |\vec{a}|^2 |\vec{d}|^2 - (\vec{a} \cdot \vec{d})^2 = 6 \times 24 - 16 = 144 - 16 = 128 \)