Question

Equation of the plane through the mid-point of the line segment joining the points P(4, 5, -10) and Q(-1, 2, 1) and perpendicular to PQ is:

• A. \( \vec{r} \cdot \left(\frac{3}{2}\hat{i} + \frac{7}{2}\hat{j} - \frac{9}{2}\hat{k}\right) = 45 \)
• B. \( \vec{r} \cdot (-\hat{i} + 2\hat{j} + \hat{k}) = \frac{135}{2} \)
• C. \( \vec{r} \cdot (5\hat{i} + 3\hat{j} - 11\hat{k}) + \frac{135}{2} = 0 \)
• D. \( \vec{r} \cdot (4\hat{i} + 5\hat{j} - 10\hat{k}) = 85 \)
• E. \( \vec{r} \cdot (5\hat{i} + 3\hat{j} - 11\hat{k}) = \frac{135}{2} \)

Hint:

The normal vector of a plane determines the coefficients of $x, y,$ and $z$. If the plane is perpendicular to $PQ$, the coefficients must be proportional to the differences in coordinates of $P$ and $Q$.

Answer 1

Answer: (E)

Concept: The equation of a plane passing through a point \( \vec{a} \) and perpendicular to a normal vector \( \vec{n} \) is given by \( (\vec{r} - \vec{a}) \cdot \vec{n} = 0 \), or \( \vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n} \). Here, the normal vector \( \vec{n} \) is the vector \( \vec{PQ} \), and the point \( \vec{a} \) is the midpoint of \( PQ \).

Step 1: Find the normal vector \( \vec{n} = \vec{PQ} \).
\( P = (4, 5, -10), Q = (-1, 2, 1) \). \[ \vec{n} = \vec{PQ} = (-1 - 4)\hat{i} + (2 - 5)\hat{j} + (1 - (-10))\hat{k} \] \[ \vec{n} = -5\hat{i} - 3\hat{j} + 11\hat{k} \] To simplify, we can use \( \vec{n} = 5\hat{i} + 3\hat{j} - 11\hat{k} \) (multiplying by -1).

Step 2: Find the midpoint \( M \).
\[ M = \left( \frac{4 + (-1)}{2}, \frac{5 + 2}{2}, \frac{-10 + 1}{2} \right) = \left( \frac{3}{2}, \frac{7}{2}, -\frac{9}{2} \right) \]

Step 3: Form the plane equation \( \vec{r} \cdot \vec{n} = \vec{M} \cdot \vec{n} \).
\[ \vec{r} \cdot (5\hat{i} + 3\hat{j} - 11\hat{k}) = \left(\frac{3}{2}\right)(5) + \left(\frac{7}{2}\right)(3) + \left(-\frac{9}{2}\right)(-11) \] \[ \vec{r} \cdot (5\hat{i} + 3\hat{j} - 11\hat{k}) = \frac{15}{2} + \frac{21}{2} + \frac{99}{2} \] \[ \vec{r} \cdot (5\hat{i} + 3\hat{j} - 11\hat{k}) = \frac{15 + 21 + 99}{2} = \frac{135}{2} \]