Question

If the position vectors of the points A and B are 3$\hat {i}$ + $\hat {j}$ + 2$\hat {k}$ and $\hat {i}$ -2$\hat {j}$ -4$\hat {k}$ respectively, then the equation of the plane through B and perpendicular to AB is

• A. $2x + 3y + 6z + 28 = 0$
• B. $2x + 3y + 6z – 11 = 0$
• C. $2x – 3y – 6z – 32 = 0$
• D. $2x + 3y + 6z + 9 = 0$

Answer 1

Answer: (A)

$2x + 3y + 6z + 28 = 0$

Answer 2

Given:
Position vector of point A: $\vec{r}_A = 3\hat{i} + \hat{j} + 2\hat{k}$
Position vector of point B: $\vec{r}_B = \hat{i} - 2\hat{j} - 4\hat{k}$

Find Vector $\vec{AB}$:
$\vec{AB} = \vec{r}_B - \vec{r}_A$
$\vec{AB} = (\hat{i} - 2\hat{j} - 4\hat{k}) - (3\hat{i} + \hat{j} + 2\hat{k})$
$\vec{AB} = -2\hat{i} - 3\hat{j} - 6\hat{k}$

Equation of the Plane:
The equation of the plane perpendicular to $\vec{AB}$ passing through point B $(1, -2, -4)$ is given by:
$2x + 3y + 6z = 2 \cdot 1 + 3 \cdot (-2) + 6 \cdot (-4)$
$2x + 3y + 6z = 2 - 6 - 24$
$2x + 3y + 6z = -28$
$2x + 3y + 6z + 28=0$

So, the correct option is (A): $2x + 3y + 6z +28=0$