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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