Question

Find the vector equation of the plane passing through the point \( A(-1, 2, -5) \) and parallel to the vectors \( 4\hat{i} - \hat{j} + 3\hat{k} \) and \( \hat{i} + \hat{j} - \hat{k} \).

Hint:

Plane's normal is the cross product of two parallel vectors; use \( \vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n} \) for vector equation.

Answer 1

The vector equation of a plane through point \( \vec{a} \) with normal perpendicular to vectors \( \vec{b} \) and \( \vec{c} \) is \( [\vec{r} - \vec{a}, \vec{b}, \vec{c}] = 0 \). 

Point \( A(-1, 2, -5) \): \( \vec{a} = -\hat{i} + 2\hat{j} - 5\hat{k} \). 

Vectors: \( \vec{b} = 4\hat{i} - \hat{j} + 3\hat{k} \), \( \vec{c} = \hat{i} + \hat{j} - \hat{k} \). 

Normal vector: \( \vec{b} \times \vec{c} \). 

\[ \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -1 & 3 \\ 1 & 1 & -1 \end{vmatrix} \\ = \hat{i} [(-1)(-1) - (3)(1)] - \hat{j} [(4)(-1) - (3)(1)] + \hat{k} [(4)(1) - (-1)(1)] \\ = \hat{i} (1 - 3) - \hat{j} (-4 - 3) + \hat{k} (4 + 1) \\ = -2\hat{i} + 7\hat{j} + 5\hat{k}. \] 

Plane equation: \( [(\vec{r} - \vec{a}), \vec{b}, \vec{c}] = 0 \). 

Or, normal form: \( (\vec{r} - \vec{a}) \cdot (\vec{b} \times \vec{c}) = 0 \). 

\[ \vec{r} = x\hat{i} + y\hat{j} + z\hat{k}, \vec{r} - \vec{a} = (x + 1)\hat{i} + (y - 2)\hat{j} + (z + 5)\hat{k}. \] \[ (x + 1)(-2) + (y - 2)(7) + (z + 5)(5) = 0 \Rightarrow -2x - 2 + 7y - 14 + 5z + 25 = 0 \Rightarrow -2x + 7y + 5z + 9 = 0. \] 

Vector equation: \( \vec{r} \cdot (-2\hat{i} + 7\hat{j} + 5\hat{k}) + 9 = 0 \). 

Answer: \( \vec{r} \cdot (-2\hat{i} + 7\hat{j} + 5\hat{k}) + 9 = 0 \).