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Three planes have the equation $$ \pi_1:x+2y-3z=-4$$ $$ \pi_2:2x+y-4z=3$$ $$ \pi_3:x+2y-3z+4+\lambda(2x+y-4z-3)=0$$

Find the equation of the plane $\pi_3$ which passes through the line on which $\pi_1$ and $\pi_2$ meet and contains the point whose position vector is $\alpha k$

I first calculated the vector equation of the line on which the two planes meet. Using the position vector of the line and the $\alpha k$, I calculated a vector that $\pi_3$ contains. Cross product of line direction vector and obtained vector gives the normal to the plane. The dot product of the normal and the $\alpha k$ position vector completes the equation of the plane.

I obtained $$-(2\alpha +5)x+(5\alpha +2)y+7z=7a$$

The given answer is $$(-2\alpha +11)x+(5\alpha +10)y-25z=-25a$$

Can somebody please check if these answers are same? If no, then maybe explain the flaw in my strategy?

P.S. I am aware of the second method that involves directly putting the position vector in the $\pi_3$ plane.

Working

Line equation:$2/3 i +5/3 j+(5i+2j+3k)t$

Normal Obtained: $(5,2,3)\times(2/3,5/3,-\alpha)=-(2\alpha +5)i+(5\alpha +2)j+7k$

Dot product of normal and position vector :$7a$

If some answers seems doubtful, upon request, I will provide detailed working.

mathnoob123
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  • The “correct” answer that you’ve given here doesn’t look right. I get $-25\alpha$ after substituting into the given equation for $\pi_3$. – amd May 23 '17 at 07:11
  • You appear to have made a mistake in calculating $\pi_1\cap\pi_2$. No point of the line that you claim is this intersection lies in $\pi_1$. – amd May 23 '17 at 07:22
  • Yes got it thanks – mathnoob123 May 23 '17 at 10:52

0 Answers0