Question

vector addition and subtraction
quiz complete
vector u has initial point at (3,5) and terminal point at (-7,5). vector v has initial point at (1,-4) and terminal
point at (6,-3).
what is u + v in component form?
(10,-4)
(5,-1)
(3,9)
(5,3)

Explanation:

Step1: Find the component form of vector \(u\)

If a vector has initial point \((x_1,y_1)\) and terminal point \((x_2,y_2)\), its component form is \(\langle x_2 - x_1,y_2 - y_1
angle\).
For vector \(u\) with initial point \((3,5)\) and terminal point \((-7,5)\), we have \(u=\langle-7 - 3,5 - 5
angle=\langle- 10,0
angle\).

Step2: Find the component form of vector \(v\)

For vector \(v\) with initial point \((1,-4)\) and terminal point \((6,-3)\), we have \(v=\langle6 - 1,-3+4
angle=\langle5,1
angle\).

Step3: Calculate \(u + v\)

If \(u=\langle a_1,b_1
angle\) and \(v=\langle a_2,b_2
angle\), then \(u + v=\langle a_1+a_2,b_1 + b_2
angle\).
Here \(a_1=-10,b_1 = 0,a_2=5,b_2 = 1\), so \(u + v=\langle-10 + 5,0+1
angle=\langle-5,1
angle\).

Answer:

\(\langle-5,1
angle\) (It seems there is a mistake in the given options as the correct result of \(u + v\) is \(\langle-5,1
angle\) which is not among the provided choices. If we assume it is \(u - v\):

Step1: Calculate \(u - v\)

If \(u=\langle a_1,b_1
angle\) and \(v=\langle a_2,b_2
angle\), then \(u - v=\langle a_1 - a_2,b_1 - b_2
angle\).
Here \(a_1=-10,b_1 = 0,a_2=5,b_2 = 1\), so \(u - v=\langle-10-5,0 - 1
angle=\langle-15,-1
angle\). If it is \(v - u\):

Step1: Calculate \(v - u\)

\(v - u=\langle5+10,1 - 0
angle=\langle15,1
angle\). If we consider calculation errors in the problem - writing (maybe wrong initial/terminal points assumed in a different way for a “standard” answer among options):
If we assume for \(u\): \(x\) - component: \(-7-3=-10\), \(y\) - component: \(5 - 5 = 0\); for \(v\): \(x\) - component: \(6 - 1=5\), \(y\) - component: \(-3+4 = 1\). And if there is a mis - take in sign in the problem setup (e.g. if \(u\) was supposed to have terminal point \((7,5)\) (typo in the problem)), \(u=\langle7 - 3,5 - 5
angle=\langle4,0
angle\), \(u + v=\langle4 + 5,0+1
angle=\langle9,1
angle\) (not in options). If \(u\) has terminal point \((- 3,5)\), \(u=\langle-3-3,5 - 5
angle=\langle-6,0
angle\), \(u + v=\langle-6 + 5,0+1
angle=\langle-1,1
angle\) (not in options). If we consider another approach:
The formula for a vector \(\vec{A}\) with initial \((x_i,y_i)\) and terminal \((x_t,y_t)\) is \(\vec{A}=(x_t - x_i)\hat{i}+(y_t - y_i)\hat{j}\)
For \(u\): \((-7-3)\hat{i}+(5 - 5)\hat{j}=-10\hat{i}+0\hat{j}\)
For \(v\): \((6 - 1)\hat{i}+(-3 + 4)\hat{j}=5\hat{i}+1\hat{j}\)
\(u + v=(-10 + 5)\hat{i}+(0 + 1)\hat{j}=-5\hat{i}+\hat{j}=\langle-5,1
angle\). If we assume the problem was \(u + v\) with wrong sign in \(u\)’s \(x\) - component calculation (maybe \(3-(-7)\) instead of \(-7 - 3\)): \(u=\langle10,0
angle\), \(u + v=\langle10+5,0 + 1
angle=\langle15,1
angle\) (not in options). If \(u\)’s \(y\) - component was miscalculated as \(5-5 = 0\) (correct) and \(v\)’s \(y\) - component: if terminal \(y=-3\) and initial \(y=-4\), \(y\) - component of \(v\) is \(-3+4 = 1\) (correct). If we assume the problem is \(u + v\) with a typo in options and we consider the closest in terms of \(x\) - component magnitude:
If we made a mistake in signs for \(u\)’s \(x\) - component (took \(3-(-7)=10\) instead of \(-7 - 3=-10\)), \(u + v=\langle10+5,0 + 1
angle=\langle15,1
angle\) (not in options). If we consider \(u - v\) with wrong \(u\) calculation (\(u=\langle-7+3,5 - 5
angle=\langle-4,0
angle\) (wrong formula application), \(u - v=\langle-4-5,0 - 1
angle=\langle-9,-1
angle\) (not in options). If \(u=\langle-7-3,5 - 5
angle=\langle-10,0
angle\) and \(v=\langle6 - 1,-3 - (-4)
angle=\langle5,1
angle\) and assume the problem was \(v - u\): \(v - u=\langle5+10,1-0
angle=\langle15,1
angle\) (not in options). If we consider a wrong formula application \((x_i - x_t,y_i - y_t)\) for vectors:
For \(u\): \((3 + 7,5 - 5)=\langle10,0
angle\), for \(v\): \((1 - 6,-4+3)=\langle-5,-1
angle\), \(u + v=\langle10-5,0-1
angle=\langle5,-1
angle\) which is the second option. So if the formula \((x_i - x_t,y_i - y_t)\) (reverse of the correct formula \((x_t - x_i,y_t - y_i)\) was used by mistake, the answer is \((5,-1)\))