Question

Let \(\vec a=\hat i+\hat j−\hat k\) and \(\vec c=2\hat i−3\hat j+2\hat k\). Then the number of vectors \(\vec b\) such that \(\vec b×\vec c=\vec a\) and \(|\vec b|∈{1,2,…,10}\) is :

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

The Correct Option is A

To solve for the number of vectors \(\vec{b}\) such that \(\vec{b} \times \vec{c} = \vec{a}\), where \(\vec{a} = \hat{i} + \hat{j} - \hat{k}\) and \(\vec{c} = 2\hat{i} - 3\hat{j} + 2\hat{k}\), and \(|\vec{b}| \in \{1, 2, \ldots, 10\}\), let's proceed step-by-step.

1. First, recall the vector cross product properties. Given vectors \(\vec{u}\) and \(\vec{v}\), their cross product \(\vec{u} \times \vec{v}\) is a vector perpendicular to both \(\vec{u}\) and \(\vec{v}\).
2. This means that the cross product \(\vec{b} \times \vec{c}\) must be parallel to \(\vec{a}\). Hence, a necessary condition is for \(\vec{a} \cdot \vec{c} = 0\), since \(\vec{a}\) must be perpendicular to \(\vec{c}\).
3. Calculate \(\vec{a} \cdot \vec{c}\): \[ \vec{a} \cdot \vec{c} = (1)(2) + (1)(-3) + (-1)(2) = 2 - 3 - 2 = -3 \] Since \(\vec{a} \cdot \vec{c} \neq 0\), \(\vec{a}\) is not perpendicular to \(\vec{c}\).
4. The condition for \(\vec{b} \times \vec{c} = \vec{a}\) cannot be satisfied because \(\vec{a}\) should be perpendicular to \(\vec{c}\) according to the nature of the cross product.
5. Thus, there are no vectors \(\vec{b}\) that satisfy both \(\vec{b} \times \vec{c} = \vec{a}\) and \(|\vec{b}| \in \{1, 2, \ldots, 10\}\).

Therefore, the number of such vectors \(\vec{b}\) is 0.