3D Geometry
Equation of a Plane
Grade 12
Question:
<p>The vector equation of the plane through the point \(\vec{i} + 2\vec{j} - \vec{k}\) and perpendicular to the line of intersection of the planes \(\vec{r} \cdot (3\vec{i} - \vec{j} + \vec{k}) = 1\) and \(\vec{r} \cdot (\vec{i} + 4\vec{j} - 2\vec{k}) = 2\), is</p>
<p>(a) \(\vec{r} \cdot (2\vec{i} + 7\vec{j} - 13\vec{k}) = 1\)</p>
<p>(b) \(\vec{r} \cdot (2\vec{i} - 7\vec{j} - 13\vec{k}) = 1\)</p>
<p>(c) \(\vec{r} \cdot (2\vec{i} + 7\vec{j} + 13\vec{k}) = 0\)</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: The direction of the line of intersection is found using the cross product of the normals of the two planes. A plane perpendicular to this line has a normal parallel to this direction vector.
Step 1: The line of intersection of the two planes is common to both planes, so it is perpendicular to the normals of both planes: \(\vec{n}_1 = 3\hat{i} - \hat{j} + \hat{k}\) and \(\vec{n}_2 = \hat{i} + 4\hat{j} - 2\hat{k}\). Step 2: The direction vector of the line of intersection is parallel to \(\vec{n}_1 \times \vec{n}_2 = -2\hat{i} + 7\hat{j} + 13\hat{k}\). Step 3: Since the required plane is perpendicular to this line of intersection, the normal to the plane is parallel to \(\vec{n}_1 \times \vec{n}_2\). Step 4: The equation of the plane with normal \(\vec{n} = 2\hat{i} - 7\hat{j} - 13\hat{k}\) passing through \(\hat{i} + 2\hat{j} - \hat{k}\) is \(\vec{r} \cdot (2\hat{i} - 7\hat{j} - 13\hat{k}) = 1\). ∴ Answer is (b).
Correct Answer: B