Question:
The mirror image of the point \(\mathrm{P}(-1,2,-4)\) in the plane \(x-y-2 z+1=0\) is
The mirror image of the point \(\mathrm{P}(-1,2,-4)\) in the plane \(x-y-2 z+1=0\) is
Show Hint
Use the image formula: image = P minus 2 times (value of plane at P over sum of squares of coefficients) times the normal vector.
Updated On: Nov 13, 2026
\((3,-4,1)\)
\((-3,4,0)\)
\((4,1,0)\)
\((2,-3,0)\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
For the plane \(ax+by+cz+d=0\), the image of \(P(x_1,y_1,z_1)\) is \(P-\frac{2(ax_1+by_1+cz_1+d)}{a^2+b^2+c^2}(a,b,c)\).
Step 2: Evaluate the plane at P
\(x-y-2z+1=-1-2+8+1=6\) and \(a^2+b^2+c^2=1+1+4=6\).
Step 3: Calculate the image
Image \(=(-1,2,-4)-\frac{2\times6}{6}(1,-1,-2)=(-1-2,\ 2+2,\ -4+4)=(-3,4,0)\).
Step 4: Compare the options
This is option (B).
Final Answer:
The mirror image is \((-3,4,0)\). \[ \boxed{\text{(B) }(-3,4,0)} \]
For the plane \(ax+by+cz+d=0\), the image of \(P(x_1,y_1,z_1)\) is \(P-\frac{2(ax_1+by_1+cz_1+d)}{a^2+b^2+c^2}(a,b,c)\).
Step 2: Evaluate the plane at P
\(x-y-2z+1=-1-2+8+1=6\) and \(a^2+b^2+c^2=1+1+4=6\).
Step 3: Calculate the image
Image \(=(-1,2,-4)-\frac{2\times6}{6}(1,-1,-2)=(-1-2,\ 2+2,\ -4+4)=(-3,4,0)\).
Step 4: Compare the options
This is option (B).
Final Answer:
The mirror image is \((-3,4,0)\). \[ \boxed{\text{(B) }(-3,4,0)} \]
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