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

The Correct Option is A

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): \(2x + 3y + 6z +28=0\)

Answer 2

Given that, the position vector of point A =3\(\hat {i}\) + \(\hat {j}\) + 2\(\hat {k}\),

and position vector of a point \(B\)= \(\hat {i}\) - 2\(\hat {j}\) - 4\(\hat {k}\)

then, \(\overrightarrow {AB}\) = Position vector of \(B\) \(\)- Position vector of \(A\)

=  \(\hat {i}\) - 2\(\hat {j}\) - 4\(\hat {k}\)- 3\(\hat {i}\) - \(\hat {j}\) - 2\(\hat {k}\)

=-\(2\)\(\hat {i}\)-\(3\hat {j}\)-\(6\)\(\hat {k}\)

Equation of the plane passing through point \(B\)  and perpendicular to \(AB\) is 

\(n.r\) =(-\(2\hat {i}\) - \(3\hat {j}\) - \(6\hat {k}\)) \(\cdot\) (\(x\hat {i}\) + \(y\hat {j}\) + \(z\hat {k}\)) 

= (-\(2\hat {i}\) - \(3\hat {j}\) - \(6\hat {k}\)) \(\cdot\) (\(\hat {i}\) - \(2\hat {j}\) - \(4\hat {k}\)),
\(⇒2x + 3y + 6z = 2 - 6 - 24\)
\(⇒2x - 2 + 3y + 6 + 6z + 24 = 0\)
\(⇒2x + 3y + 6z + 28 = 0\)  (Ans.)