Question

Find the orthogonal projection of a point A = $1, 2, -1$ onto a line passing through the points Pi = $0, 1, 1$ and P2 = $1, 2, 3$.

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Answer to a math question Find the orthogonal projection of a point A = $1, 2, -1$ onto a line passing through the points Pi = $0, 1, 1$ and P2 = $1, 2, 3$.

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To find the orthogonal projection of a point A onto a line passing through the points P1 and P2, we need to determine the closest point on the line to A. Let's call this point P.

Step 1: Find the direction vector of the line
To find the direction vector of the line passing through P1 and P2, we subtract the coordinates of P1 from P2:

\vec{v} = \vec{P2} - \vec{P1} = $1, 2, 3$ - $0, 1, 1$ = $1, 1, 2$

Step 2: Find the vector from P1 to A
To find the vector from P1 to A, we subtract the coordinates of P1 from the coordinates of A:

\vec{u} = \vec{A} - \vec{P1} = $1, 2, -1$ - $0, 1, 1$ = $1, 1, -2$

Step 3: Find the projection of vector u onto vector v
The orthogonal projection of u onto v is given by the formula:

proj_v$u$ = \frac{\vec{u} \cdot \vec{v}}{\|\vec{v}\|^2} \cdot \vec{v} = \frac{$1, 1, -2$ \cdot $1, 1, 2$}{\|$1, 1, 2$\|^2} \cdot $1, 1, 2$

Step 4: Calculate the dot product and norm
The dot product of vectors u and v is:

\vec{u} \cdot \vec{v} = 1 \cdot 1 + 1 \cdot 1 + $-2$ \cdot 2 = -2

The norm of vector v is:

\|\vec{v}\| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{6}

Step 5: Calculate the projection vector
Substituting the values into the formula, we have:

proj_v$u$ = \frac{-2}{6} \cdot $1, 1, 2$ = \left$-\frac{1}{3}, -\frac{1}{3}, -\frac{2}{3}\right$

Step 6: Find the point P on the line passing through P1 and P2
To find the point P, we add the coordinates of P1 to the projection vector:

\vec{P} = \vec{P1} + proj_v$u$ = $0, 1, 1$ + \left$-\frac{1}{3}, -\frac{1}{3}, -\frac{2}{3}\right$ = \left$-\frac{1}{3}, \frac{2}{3}, \frac{1}{3}\right$

Answer: The orthogonal projection of point A = $1, 2, -1$ onto the line passing through Pi = $0, 1, 1$ and P2 = $1, 2, 3$ is point P = \left$-\frac{1}{3}, \frac{2}{3}, \frac{1}{3}\right$.

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