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: Cartesian form first:
A plane perpendicular to $(-5,-3,11)$ is $-5x - 3y + 11z = k$.

Step 2: Use the midpoint:
At $M\left(\frac32,\frac72,-\frac92\right)$: $k = -\frac{15}{2} - \frac{21}{2} - \frac{99}{2} = -\frac{135}{2}$.

Step 3: Vector form:
$\bar r\cdot(-5\hat i - 3\hat j + 11\hat k) = -\frac{135}{2}$, which is option (A).

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