Question

If M denotes the midpoint of the line joining A\((4,5,-10)\) and B\((-1,2,1)\), then the equation of the plane through M and perpendicular to AB is:

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

Hint:

AB gives the normal vector; plug in the midpoint to find the constant.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A plane perpendicular to \(AB\) through \(M\) has normal \(\overrightarrow{AB}\) and passes through \(M\).

Step 2: Find M and the normal:
\(M = \left(\frac{4-1}{2}, \frac{5+2}{2}, \frac{-10+1}{2}\right) = \left(\frac32, \frac72, -\frac92\right)\).
Normal \(\overrightarrow{AB} = B - A = (-5, -3, 11)\).

Step 3: Equation:
\(\bar r\cdot\bar n = \bar m\cdot\bar n\) with \(\bar m\cdot\bar n = \frac32(-5) + \frac72(-3) + \left(-\frac92\right)(11) = -\frac{15}{2} - \frac{21}{2} - \frac{99}{2} = -\frac{135}{2}\).
\[ \bar r\cdot(-5\hat i - 3\hat j + 11\hat k) + \frac{135}{2} = 0 \]

Step 4: Why the other options are wrong.
Option (B) uses the midpoint as the normal. Option (C) uses \(\overrightarrow{OA}\) as the normal. Option (D) uses the vector \(-\hat i + 2\hat j + \hat k\), which is not parallel to \(AB\).

Final Answer:
Option (A) is the plane. \[ \boxed{\bar r\cdot(-5\hat i-3\hat j+11\hat k)+\frac{135}{2}=0} \]