Question:
If the mirror image of the point \((1,3,5)\) with respect to the plane \(4x-5y+2z=8\) is \((\alpha,\beta,\gamma)\), then \(5(\alpha+\beta+\gamma)\) equals :
If the mirror image of the point \((1,3,5)\) with respect to the plane \(4x-5y+2z=8\) is \((\alpha,\beta,\gamma)\), then \(5(\alpha+\beta+\gamma)\) equals :
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Image of \(P\) in plane \(ax+by+cz+d=0\) is \(P-\frac{2(ax_1+by_1+cz_1+d)}{a^2+b^2+c^2}(a,b,c)\).
Updated On: Nov 4, 2026
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- 47
- 43
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The Correct Option is C
Solution and Explanation
Plane: \(4x-5y+2z-8=0\), normal \((4,-5,2)\), \(a^2+b^2+c^2=16+25+4=45\).
Value at \(P(1,3,5)\): \(4-15+10-8=-9\).
Image \(=P-\frac{2(-9)}{45}(4,-5,2)=(1,3,5)+\frac25(4,-5,2)\).
\(\alpha=1+\frac85=\frac{13}{5}\), \(\beta=3-2=1\), \(\gamma=5+\frac45=\frac{29}{5}\).
\(\alpha+\beta+\gamma=\frac{13+5+29}{5}=\frac{47}{5}\), so \(5(\alpha+\beta+\gamma)=47\).
Answer: (3) 47
Value at \(P(1,3,5)\): \(4-15+10-8=-9\).
Image \(=P-\frac{2(-9)}{45}(4,-5,2)=(1,3,5)+\frac25(4,-5,2)\).
\(\alpha=1+\frac85=\frac{13}{5}\), \(\beta=3-2=1\), \(\gamma=5+\frac45=\frac{29}{5}\).
\(\alpha+\beta+\gamma=\frac{13+5+29}{5}=\frac{47}{5}\), so \(5(\alpha+\beta+\gamma)=47\).
Answer: (3) 47
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