Question

If the product of the distances of the point (1, 1, 1) from the origin and the plane x − y + z+ λ = 0 be 5, find the value of λ.

Answer 1

\[ \text{ We know that the distance of the point } \left( x_1 , y_1 , z_1 \right) \text{ from the plane  ax + by + cz + d = 0 is given by } \]

\[\frac{\left| a x_1 + b y_1 + c z_1 + d \right|}{\sqrt{a^2 + b^2 + c^2}}\]

\[ \text{ Distance of the point (1, 1, 1) from the plane }  x-y+z+\lambda=0\]

\[\text{ The required distance} \]

\[ = \frac{\left| 1 - 1 + 1 + \lambda \right|}{\sqrt{1^2 + \left( - 1 \right)^2 + 1^2}}\]

\[ = \frac{\left| 1 + \lambda \right|}{\sqrt{3}} \text{ units }  ... (1)\]

\[ \text{ Distance of the point (0, 0, 0) from the plane } x-y+z+\lambda=0\]

\[ \text{ The required distance } \]

\[ = \frac{\left| 0 - 0 + 0 + \lambda \right|}{\sqrt{1^2 + \left( - 1 \right)^2 + 1^2}}\]

\[ = \frac{\left| \lambda \right|}{\sqrt{3}} \text{ units }  ... (2)\]

\[\text{ It is given that the product of the distances (1) and (2) is 5 } .\]

\[ \Rightarrow \frac{\left| 1 + \lambda \right|}{\sqrt{3}} \times \frac{\left| \lambda \right|}{\sqrt{3}} = 5\]

\[ \Rightarrow \lambda^2 + \lambda - 15 = 0\]