Using image formula: $\frac{x-1}{1} = \frac{y-2}{1} = \frac{z-3}{1} = \frac{-2(1+2+3-9)}{1^2+1^2+1^2} = \frac{6}{3} = 2$
Image $= (1+2, 2+2, 3+2) = (3, 4, 5)$
Verified 30 May 2026.
The image of the point $(1, 2, 3)$ in the plane $x + y + z = 9$ is:
$(3, 4, 5)$
$(5, 4, 3)$
$(2, 3, 4)$
$(3, 3, 3)$
Using image formula: $\frac{x-1}{1} = \frac{y-2}{1} = \frac{z-3}{1} = \frac{-2(1+2+3-9)}{1^2+1^2+1^2} = \frac{6}{3} = 2$
Image $= (1+2, 2+2, 3+2) = (3, 4, 5)$
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The volume of the parallelepiped formed by vectors a=i+2j-k, b=2i-j+3k, c=3i+j+2k is:
The position vectors of points A and B are 2i+3j+k and 4i+j-2k. The position vector of the midpoint of AB is:
The projection of vector a=2i-3j+6k on vector b=i+2j+2k is:
The scalar triple product [i j k] is:
The equation of the plane passing through (1,2,3) and perpendicular to the vector 2i+3j-k is: