3D Geometry
Plane equations
Grade 12
Question:
<p>The vector equation of plane which is at a distance of 8 units from the origin and which is normal to the vector \(2\vec{i} + \vec{j} + 2\vec{k}\) is \(\vec{r} \times (2\vec{i} + \vec{j} + 2\vec{k}) = l\), where \(l\) is equal to</p>
<p>(a) 0</p>
<p>(b) 24</p>
<p>(c) 42</p>
<p>(d) 8</p>
Step-by-Step Solution
Key Concept: The constant in the vector equation of a plane equals the distance from origin times the magnitude of the normal vector.
Step 1: Given distance $d = 8$ and normal vector $\vec{n} = 2\hat{i} + \hat{j} + 2\hat{k}$. Step 2: The magnitude of the normal vector is $|\vec{n}| = \sqrt{2^2 + 1^2 + 2^2} = \sqrt{4+1+4} = \sqrt{9} = 3$. Step 3: For a plane at distance $d$ from origin with normal vector $\vec{n}$, the vector equation is $\vec{r} \cdot \vec{n} = d|\vec{n}|$. Step 4: Therefore, $l = d \times |\vec{n}| = 8 \times 3 = 24$. ∴ Answer is (b) 24.
Correct Answer: B