The equation of the plane which bisects the line segment joining (2, 3, 4) and (6, 7, 8), is
Answer & explanation
Correct answer: option 4
Let the two points be P(2, 3, 4) and Q(6, 7, 8) and let R be their mid-point. The required plane passes through the mid-point R and is normal to PQ. So, its equation is
$[\vec{r}. (4\hat{i} + 5\hat{j} + 6\hat{k})]. (4\hat{i} + 4\hat{j} + 4\hat{k})= 0 $
$⇒ \vec{r}. (\hat{i} + \hat{j} + \hat{k})- ( 4 + 5 + 6) = 0 ⇒ x + y + z - 15 = 0 $