Question

Find the vector equation of a plane which passes through the point of intersection of the planes \[ \vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 6 \, \text{and} \, \vec{r} \cdot (2\hat{i} + 3\hat{j} + 4\hat{k}) = -5 \] and the point $(1,1,1)$.

Hint:

The plane through the intersection of two planes always involves a parameter $\lambda$, which is determined using the given point.

Answer 1

Step 1: Equation of required plane. The equation of a plane passing through the line of intersection of two planes is: \[\vec{r} \cdot \vec{n}_1 = d_1 + \lambda (\vec{r} \cdot \vec{n}_2 - d_2)\] Here, \[\vec{n}_1 = (1,1,1), d_1 = 6\] \[\vec{n}_2 = (2,3,4), d_2 = -5\] So, \[\vec{r} \cdot (1,1,1) - 6 + \lambda \big(\vec{r} \cdot (2,3,4) + 5 \big) = 0\]

Step 2: General equation of the plane. \[(x+y+z - 6) + \lambda (2x + 3y + 4z + 5) = 0\]

Step 3: Condition of passing through $(1,1,1)$. Substitute $x=1, y=1, z=1$: \[(1+1+1 - 6) + \lambda (2(1)+3(1)+4(1)+5) = 0\] \[(-3) + \lambda (14) = 0 \implies \lambda = \frac{3}{14}\]

Step 4: Required plane. Substitute $\lambda = \frac{3}{14}$: \[(x+y+z - 6) + \frac{3}{14}(2x + 3y + 4z + 5) = 0\] Multiply through by $14$: \[14(x+y+z - 6) + 3(2x + 3y + 4z + 5) = 0\] \[14x + 14y + 14z - 84 + 6x + 9y + 12z + 15 = 0\] \[20x + 23y + 26z - 69 = 0\]

Final Answer: \[\boxed{20x + 23y + 26z - 69 = 0}\]