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Question:

Let (\vec{a} = \hat{i} - \hat{j}), (\vec{b} = \hat{i} + \hat{j} + \hat{k}) and (\vec{c}) be a vector such that (\vec{a} \times \vec{c} + \vec{b} = \vec{0}) and (\vec{a} . \vec{c} = 4), then (|\vec{c}|^2) is equal to

192

9

8

172

Solution:

(\vec{a} \times \vec{c} = -\vec{b})
Taking cross product with (\vec{a}) on both sides,
((\vec{a} \times \vec{c}) \times \vec{a} = -\vec{b} \times \vec{a})
( (\vec{c} (\vec{a} . \vec{a}) - \vec{a} (\vec{a} . \vec{c})) = \vec{a} \times \vec{b})
(2\vec{c} - 4\vec{a} = \vec{a} \times \vec{b})
(\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 0 \ 1 & 1 & 1 \end{vmatrix} = -\hat{i} - \hat{j} + 2\hat{k})
(2\vec{c} = 4\vec{a} + \vec{a} \times \vec{b} = 4(\hat{i} - \hat{j}) + (-\hat{i} - \hat{j} + 2\hat{k}) = 3\hat{i} - 5\hat{j} + 2\hat{k})
(\vec{c} = \frac{3}{2}\hat{i} - \frac{5}{2}\hat{j} + \hat{k})
(|\vec{c}|^2 = \frac{9}{4} + \frac{25}{4} + 1 = \frac{38}{4} = \frac{19}{2})
This is not matching with any option. Let's check the calculation again.
(\vec{a} \times \vec{c} = -\vec{b})
Taking dot product with (\vec{a}) on both sides,
(\vec{a} . (\vec{a} \times \vec{c}) = -\vec{a} . \vec{b})
Since (\vec{a} . (\vec{a} \times \vec{c}) = 0), (\vec{a} . \vec{b} = 0)
(\vec{a} . \vec{b} = (\hat{i} - \hat{j}) . (\hat{i} + \hat{j} + \hat{k}) = 1 - 1 = 0)
(\vec{a} \times \vec{c} = -\vec{b})
(|\vec{a} \times \vec{c}|^2 = |\vec{b}|^2)
(|\vec{a}|^2 |\vec{c}|^2 sin^2\theta = |\vec{b}|^2)
(2|\vec{c}|^2 sin^2\theta = 3)
(\vec{a} . \vec{c} = |\vec{a}| |\vec{c}| cos\theta = 4)
(\sqrt{2}|\vec{c}| cos\theta = 4)
(|\vec{c}| cos\theta = 2\sqrt{2})
(|\vec{c}|^2 cos^2\theta = 8)
(|\vec{c}|^2 (sin^2\theta + cos^2\theta) = \frac{3}{2} + 8 = \frac{19}{2})
(|\vec{c}|^2 = \frac{19}{2})