What is the projection of (2,6) onto (-1,5)?
The orthogonal projection is (( (2,6) dot (-1,5) ) / ( (-1,5) dot (-1,5) ))*(-1,5)
To find the projection of a point onto a line, you can use the concept of vectors and dot product. Here's how you can find the projection of the point (2,6) onto the line with direction vector (-1,5):
Step 1: Calculate the unit vector of the line's direction vector. The unit vector is found by dividing the direction vector by its magnitude.
Magnitude of direction vector = sqrt((-1)^2 + 5^2) = sqrt(1 + 25) = sqrt(26)
Unit vector = (-1/sqrt(26), 5/sqrt(26))
Step 2: Calculate the dot product of the vector from the line's starting point to the given point and the unit vector.
Vector from starting point to given point = (2 - (-1), 6 - 5) = (3, 1)
Dot product = (3, 1) • (-1/sqrt(26), 5/sqrt(26)) = (3 * -1/sqrt(26)) + (1 * 5/sqrt(26)) = (-3/sqrt(26)) + (5/sqrt(26)) = 2/sqrt(26)
Step 3: Multiply the dot product with the unit vector to find the projection of the point onto the line.
Projection = (2/sqrt(26)) * (-1/sqrt(26), 5/sqrt(26)) = (-2/sqrt(26), 10/sqrt(26))
Therefore, the projection of the point (2,6) onto the line with direction vector (-1,5) is approximately (-0.39, 1.96).
To find the projection of a point onto a line, we need to calculate the vector projection. The vector projection of vector A onto vector B can be computed using the formula:
Proj A onto B = (A dot B) / ||B||^2 * B
Where "." represents the dot product and "||B||" represents the magnitude of B.
First, let's calculate the dot product of vectors A and B:
A dot B = (2 * -1) + (6 * 5) = -2 + 30 = 28
Next, let's calculate the magnitude of vector B:
||B|| = √((-1)^2 + (5)^2) = √(1 + 25) = √26
Finally, substitute the values into the formula for the vector projection:
Proj A onto B = (A dot B) / ||B||^2 * B
= 28 / (√26)^2 * (-1, 5)
= 28 / 26 * (-1, 5)
= (-28/26, 140/26)
Simplifying this projection gives (-14/13, 70/13). Therefore, the projection of the point (2, 6) onto the line (-1, 5) is approximately (-1.08, 5.39).