Let U and V be the points with position vectors u = (1, 2, −1, −2) and v = (0,3,−2,2,).
Find the distance from U to the span(v
To find the distance from point U to the span of vector v, we can use the formula:
distance = ||proj_uv - u||
where proj_uv is the projection of vector u onto vector v and ||•|| denotes the norm or magnitude of a vector.
To find the projection proj_uv, we can use the formula:
proj_uv = (u · v / ||v||²) * v
where u · v is the dot product of u and v, and ||v||² is the square of the norm or magnitude of vector v.
The dot product u · v is calculated as:
(u · v) = (1*0) + (2*3) + (-1*-2) + (-2*2) = 6
The square of the norm of v, ||v||², is calculated as:
||v||² = (0^2) + (3^2) + (-2^2) + (2^2) = 17
Substituting these values into the formula for proj_uv, we find:
proj_uv = (6 / 17) * (0,3,−2,2) = (0, 6/17, -12/17, 12/17)
Now, to calculate the norm of proj_uv - u, we subtract u from proj_uv:
proj_uv - u = (0, 6/17, -12/17, 12/17) - (1, 2, -1, -2) = (-1, 6/17, -5/17, 30/17)
Finally, we calculate the norm or magnitude of proj_uv - u:
||proj_uv - u|| = sqrt((-1)^2 + (6/17)^2 + (-5/17)^2 + (30/17)^2)
= sqrt(1 + 36/289 + 25/289 + 900/289)
= sqrt(1 + 61/289 + 900/289)
= sqrt(365/289)
= sqrt(365) / sqrt(289)
= 19.1
Therefore, the distance from U to the span of vector v is approximately 19.1.