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A line has the equation $$r = 4i -j + 5k + \lambda(i - j +2k)$$ A plane has the equation $$ 3x + 2y + 4z = 12$$ The point $A$ is $(4,-1,5)$, the point $B(2, 1, 1)$ is the intersection between the plane and the line.

Find the coordinates of $C$ where $C$ lies on the plane and forms the right triangle with $A$ and $B$.

I have taken the vector product of normal vector of the plane and the direction vector of the line which gives $8x - 2y -6z$.

I am stuck as how to get the coordinates of C.

Thanks in advance for advice!

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  • $\begingroup$ Hint: you basically want the perpendicular (orthogonal) projection of the line on the plane...:) $\endgroup$ – DonAntonio Apr 26 '14 at 16:59
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Let $\mathbf{n}$ be the vector $\langle3,2,4\rangle$ normal to the plane. Note \begin{align} \mathbf{w}=\stackrel{\longrightarrow}{BA}-\text{ proy}_{\mathbf{n}}{\stackrel{\longrightarrow}{BA}}&=\stackrel{\longrightarrow}{BA}-\left(\frac{\stackrel{\longrightarrow}{BA}\cdot\mathbf{n}}{\left\|\mathbf{n}\right\|}\right)\frac{\mathbf{n}}{\left\|\mathbf{n}\right\|} \\ &=\langle2,-2,4\rangle-\left(\frac{\langle2,-2,4\rangle\cdot\langle3,2,4\rangle}{\sqrt{29}}\right)\frac{\langle3,2,4\rangle}{\sqrt{29}}\\ &=\langle2,-2,4\rangle-\left(\frac{18}{\sqrt{29}}\right)\frac{\langle3,2,4\rangle}{\sqrt{29}}\\ &=\langle2,-2,4\rangle-\langle\frac{54}{29},\frac{36}{29},\frac{72}{29}\rangle\\ &=\langle\frac{4}{29},-\frac{94}{29},\frac{44}{29}\rangle \end{align} is a vector normal to $\mathbf{n}$ (since $\mathbf{w}\cdot\mathbf{n}=0$), then $C$ is the final point of $\stackrel{\longrightarrow}{OB}+\mathbf{w}=\langle\frac{62}{29},-\frac{65}{29},\frac{73}{29}\rangle$, as a sort of comprobation we can compute $\stackrel{\longrightarrow}{AC}\cdot\stackrel{\longrightarrow}{CB}$: \begin{align} \stackrel{\longrightarrow}{AC}\cdot\stackrel{\longrightarrow}{CB} & =\langle-\frac{54}{29},-\frac{36}{29},-\frac{72}{29}\rangle\cdot\langle-\frac{4}{29},\frac{94}{29},-\frac{44}{29}\rangle \\ &=\frac{216-3384+3168}{29^2} \\ &=0. \end{align}

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  • $\begingroup$ May I ask what you mean by a triangle ABC being "rectangle in C"? $\endgroup$ – PeteUK Apr 26 '14 at 18:51
  • $\begingroup$ Sorry, I meant right triangle at $C$: In this way we obtain a point C such that triangle △ABC is right at C. But if we wish triangle ABC be right at A we must consider the line of intersection L between the plane $3x+2y+4z=12$ and the plane by A and normal AB. $\endgroup$ – Ángel Mario Gallegos Apr 26 '14 at 18:54

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