1that is the eq of a line not a plane.
Q17. Find the image of the point (1,6,3) about the "plane"(???)
x = (y-1)/2 = (z-2)/3 <- How is this a plane??
My working -:
-
UP 0 DOWN 0 0 4
4 Answers
39I took a point M which satisfies the line to be as shown. Now vector AM dot product direction vector of line (r = a + lambda b, b is direction vector) should be zero as they are perpendicular.
Found value of lambda then used midpoint formula.
Maybe lambda has been calculated wrong? If lambda = 1, the answer comes right..
39Yes che exactly. I called one of my dumbass classmates, and he started telling me off saying I don't know 3D(about eq of plane and line). I was so incensed I nearly threw the phone..he called back but I didn't answer :).
So it was a calculation mistake!!! Otherwise my working is correct...chalo thanks sandi bhai. :D