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

$$\begin{gathered} \text{We know that the distance of the point}\left(x_1,y_1,z_1\right)\text{from the plane}ax+by+cz+d=0\text{is given by} \\ \frac{\left|a{x}_1+b{y}_1+c{z}_1+d\right|}{\sqrt{a^2+b^2+c^2}} \\ \mathbf{\text{Distance of the point (1, 1, 1) from the plane}}\mathit{\text{x-y+z+λ}}\mathbf{\text{=0}} \\ \text{The required distance} \\ =\frac{\left|1-1+1+\lambda \right|}{\sqrt{1^2+{\left(-1\right)}^2+1^2}} \\ =\frac{\left|1+\lambda \right|}{\sqrt{3}}\text{units ... (1)} \\ \mathbf{\text{Distance of the point (0, 0, 0) from the plane}}\mathit{\text{x-y+z+λ}}\mathbf{\text{=0}} \\ \text{The required distance} \\ =\frac{\left|0-0+0+\lambda \right|}{\sqrt{1^2+{\left(-1\right)}^2+1^2}} \\ =\frac{\left|\lambda \right|}{\sqrt{3}}\text{units ... (2)} \\ \text{It is given that the product of the distances (1) and (2) is 5.} \\ \Rightarrow \frac{\left|1+\lambda \right|}{\sqrt{3}}\times \frac{\left|\lambda \right|}{\sqrt{3}}=5 \\ \Rightarrow {\lambda }^2+\lambda -15=0 \end{gathered}$$

Disclaimer: The answer or problem given in the text book is incorrect for this.