<p>The length of the perpendicular from the origin to the plane passing through the point \(\vec{a}\) and containing the line \(\vec{r} = \vec{b} + \lambda \vec{c}\) is</p>
<p>(a) \(\frac{[\vec{a}\,\vec{b}\,\vec{c}]}{|\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}|}\)</p>
<p>(b) \(\frac{[\vec{a}\,\vec{b}\,\vec{c}]}{|\vec{a} \times \vec{b} + \vec{b} \times \vec{c}|}\)</p>
<p>(c) \(\frac{[\vec{a}\,\vec{b}\,\vec{c}]}{|\vec{b} \times \vec{c} + \vec{c} \times \vec{a}|}\)</p>
<p>(d) \(\frac{[\vec{a}\,\vec{b}\,\vec{c}]}{|\vec{a} \times \vec{b} + \vec{c} \times \vec{a}|}\)</p>
Step-by-Step Solution
Key Concept: The normal to the plane is the cross product of two vectors in the plane: (a - b) and c. Use the distance formula from a point to a plane.
Solution: The plane passes through point \(\vec{a}\) and also through point \(\vec{b}\) (on the line), and is parallel to vector \(\vec{c}\). The normal to the plane is perpendicular to both \((\vec{a} - \vec{b})\) and \(\vec{c}\). So the normal vector is \((\vec{a} - \vec{b}) \times \vec{c} = \vec{a} \times \vec{c} - \vec{b} \times \vec{c} = \vec{a} \times \vec{c} + \vec{c} \times \vec{b}\) The equation of the plane is: \((\vec{r} - \vec{a}) \cdot [(\vec{a} - \vec{b}) \times \vec{c}] = 0\) The perpendicular distance from origin O to this plane is: \(d = \frac{|\vec{a} \cdot [(\vec{a} - \vec{b}) \times \vec{c}]|}{|(\vec{a} - \vec{b}) \times \vec{c}|} = \frac{[\vec{a}\,(\vec{a} - \vec{b})\,\vec{c}]}{|\vec{b} \times \vec{c} + \vec{c} \times \vec{a}|}\) ∴ Answer is (c).
Correct Answer: c