Question

If the point \(Q(\alpha ,\beta ,\gamma )\) is the mirror image of \(P(10,15,20)\) in the plane \(\vec{r}=\hat{k}+\lambda (-\hat{i}+\hat{j})+\mu (-\hat{i}+\hat{k})\), \(\lambda ,\mu \in \mathbb{R}\) then the value of \(3(\alpha +\beta +\gamma )\) is

• A. \(-45\)
• B. \(-120\)
• C. \(-105\)
• D. \(-129\)

Hint:

Find the normal from the two direction vectors, write the plane equation, then use the reflection formula.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
The plane passes through \((0,0,1)\) and is parallel to \(-\hat{i}+\hat{j}\) and \(-\hat{i}+\hat{k}\).

Step 2: Find the normal
\((-\hat{i}+\hat{j})\times(-\hat{i}+\hat{k})=\hat{i}(1-0)-\hat{j}(-1-0)+\hat{k}(0+1)=\hat{i}+\hat{j}+\hat{k}\).

Step 3: Plane equation
\(x+y+z=1\) since the point \((0,0,1)\) lies on it.

Step 4: Reflect P(10, 15, 20)
\(\dfrac{\alpha-10}{1}=\dfrac{\beta-15}{1}=\dfrac{\gamma-20}{1}=\dfrac{-2(10+15+20-1)}{3}=-\dfrac{88}{3}\).

Step 5: Sum
\(\alpha+\beta+\gamma=45-88=-43\), so \(3(\alpha+\beta+\gamma)=-129\).

Final Answer:
The required value is -129. \[ \boxed{\text{(D) }-129} \]