Question

Let$\alpha ,\beta ,\gamma ,\delta$ be real numbers such that ${\alpha }^2+{\beta }^2+{\gamma }^2\ne 0$ and $\alpha +\gamma =1$.Suppose the point $(3,2,-1)$ is the mirror image of the point$(1,0,-1)$ with respect to the plane $\alpha x+\beta y+\gamma z=\delta$. Then which of the following statements is/are TRUE?

• A. $\alpha +\beta =2$
• B. $\delta –\gamma =3$
• C. $\delta +\beta =4$
• D. $\alpha +\beta +\gamma =\delta$

Answer 1

Answer: (A), (B), (C)

$\delta +\beta =4$

Explanation for correct options:

Option(A):$\mathbf{\alpha }\mathbf{}\mathbf{+}\mathbf{}\mathbf{\beta }\mathbf{}\mathbf{=}\mathbf{2}$

Finding value of $\alpha +\beta$

In the given equation of plane $\alpha x+\beta y+\gamma z=\delta ....\left(i\right)$

$Q$ point is mid point of $P(3,2,-1)$ and $P\text{'}(1,0,-1)$ then

$\mathit{Q}\mathbf{=}\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2}\right)$

Coordinates of $Q$ $=\left(\frac{(1+3)}{2},\frac{(0+2)}{2},\frac{(-1-1)}{2}\right)=(2,1,-1)$

Since the point $Q$ lies on the plane $\alpha x+\beta y+\gamma z=\delta$

$\therefore \alpha \left(2\right)+\beta \left(1\right)+\gamma (-1)=\delta \dots \left(ii\right)$

So, $PP\text{'}$ is normal to the given plane

Let $\frac{\alpha }{3-1}=\frac{\beta }{2-0}=\frac{\gamma }{-1+1}=\lambda$

$\Rightarrow \alpha =2\lambda ,\beta =2\lambda ,\gamma =0.....\left(iii\right)$

Given that $\alpha +\gamma =1$

$$\begin{gathered} \Rightarrow 2\lambda +0=1 \\ \Rightarrow \lambda =\frac{1}{2}.....\left(iv\right)[from(iii)] \end{gathered}$$

$$\begin{gathered} \therefore \alpha =2\lambda \\ =2x\frac{1}{2} \\ =1 \end{gathered}$$

Similarly from $\left(ii\right)\&\left(iii\right)$ we have $\beta =2\lambda =1$ and $\gamma =0$

So,

$$\begin{gathered} \Rightarrow \alpha +\beta =1+1 \\ \Rightarrow \alpha +\beta =2 \end{gathered}$$

Hence, option A is correct.

Option(B):$\mathit{\delta }\mathbf{}\mathbf{–}\mathbf{}\mathit{\gamma }\mathbf{}\mathbf{=}\mathbf{}\mathbf{3}$:

Finding the value of $\delta –\gamma$

Putting derived values of $\alpha ,\beta ,\gamma$ into $\left(ii\right)$

$$\begin{gathered} \Rightarrow 1\left(2\right)+1\left(1\right)+0(-1)=\delta \\ \Rightarrow \delta =3....\left(v\right) \end{gathered}$$

Therefore, $\delta –\gamma =3-0=3[from(iii\left)\right]$

Hence, option B is correct.

Option(C):$\mathit{\delta }\mathbf{}\mathbf{+}\mathbf{}\mathit{\beta }\mathbf{}\mathbf{=}\mathbf{}\mathbf{4}$:

Finding the value of $\delta +\beta$

From equation $\left(iii\right)\&\left(v\right)$

$$\begin{gathered} \Rightarrow \delta +\beta =3+1 \\ \Rightarrow \delta +\beta =4...\left(vi\right) \end{gathered}$$

Hence, option C is correct .

Explanation For The Incorrect option:

Option(D):$\mathit{\alpha }\mathbf{}\mathbf{+}\mathbf{}\mathit{\beta }\mathbf{}\mathbf{+}\mathbf{}\mathit{\gamma }\mathbf{}\mathbf{=}\mathbf{}\mathit{\delta }$

Finding the value of $\alpha +\beta +\gamma$

From above calculations we have

$\alpha =1,\beta =1,\gamma =0\&\delta =3$

$$\begin{gathered} \therefore \alpha +\beta +\gamma =1+1+0 \\ =2(\ne \delta ) \end{gathered}$$

Hence, option D is incorrect.

Hence, the correct options are A,B and C