A displacement vector shows the change in position from one point to another. To find a displacement vector from an initial point to a terminal point, subtract the coordinates of the initial point from the coordinates of the terminal point. These problems include 2D displacement vectors, 3D displacement vectors, finding initial points, finding terminal points, and using vector notation with \(\vec{i},\vec{j},\vec{k}\).
Notes

Practice Problems
\(\textbf{1)}\) Find the displacement vector between \(A (-5,8) \) and \( B (6,0)\)
\(\vec{AB}\) is \(\langle 11,-8\rangle\) or \(11\vec{i}-8\vec{j}\)
\(\vec{AB}=\langle x_2-x_1,y_2-y_1\rangle\)
\(\vec{AB}=\langle6-(-5),0-8\rangle\)
\(\vec{AB}=\langle11,-8\rangle\)
\(\vec{AB}=11\vec{i}-8\vec{j}\)
\(\textbf{2)}\) Find the displacement vector between \(A (4,2) \) and \( B (1,9)\)
\(\vec{AB}\) is \(\langle -3,7\rangle\) or \(-3\vec{i}+7\vec{j}\)
\(\vec{AB}=\langle x_2-x_1,y_2-y_1\rangle\)
\(\vec{AB}=\langle1-4,9-2\rangle\)
\(\vec{AB}=\langle-3,7\rangle\)
\(\vec{AB}=-3\vec{i}+7\vec{j}\)
\(\textbf{3)}\) Find the displacement vector between \(A (0,0) \) and \( B (1,5)\)
\(\vec{AB}\) is \(\langle 1,5\rangle\) or \(1\vec{i}+5\vec{j}\)
\(\vec{AB}=\langle x_2-x_1,y_2-y_1\rangle\)
\(\vec{AB}=\langle1-0,5-0\rangle\)
\(\vec{AB}=\langle1,5\rangle\)
\(\vec{AB}=1\vec{i}+5\vec{j}\)
\(\textbf{4)}\) Find the displacement vector between \(C (1,3,2) \) and \( D (4,-1,-5)\)
\(\vec{CD}\) is \(\langle3,-4,-7\rangle\) or \(3\vec{i}-4\vec{j}-7\vec{k}\)
\(\vec{CD}=\langle x_2-x_1,y_2-y_1,z_2-z_1\rangle\)
\(\vec{CD}=\langle4-1,-1-3,-5-2\rangle\)
\(\vec{CD}=\langle3,-4,-7\rangle\)
\(\vec{CD}=3\vec{i}-4\vec{j}-7\vec{k}\)
\(\textbf{5)}\) Find the displacement vector between \(C (5,1,6) \) and \( D (-3,-2,-1)\)
\(\vec{CD}\) is \(\langle-8,-3,-7\rangle\) or \(-8\vec{i}-3\vec{j}-7\vec{k}\)
\(\vec{CD}=\langle x_2-x_1,y_2-y_1,z_2-z_1\rangle\)
\(\vec{CD}=\langle-3-5,-2-1,-1-6\rangle\)
\(\vec{CD}=\langle-8,-3,-7\rangle\)
\(\vec{CD}=-8\vec{i}-3\vec{j}-7\vec{k}\)
\(\textbf{6)}\) Find the displacement vector between \(C (1,2,3) \) and \( D (4,-5,-6)\)
\(\vec{CD}\) is \(\langle3,-7,-9\rangle\) or \(3\vec{i}-7\vec{j}-9\vec{k}\)
\(\vec{CD}=\langle x_2-x_1,y_2-y_1,z_2-z_1\rangle\)
\(\vec{CD}=\langle4-1,-5-2,-6-3\rangle\)
\(\vec{CD}=\langle3,-7,-9\rangle\)
\(\vec{CD}=3\vec{i}-7\vec{j}-9\vec{k}\)
\(\textbf{7)}\) Find the initial point \((P)\) of vector \(\vec{v}\) from \(P\) to \(Q\) where \(\vec{v}= \langle3,8\rangle \) and \( Q= (4,1)\)
The initial point is \((1,-7)\)
\(\vec{PQ}=\langle x_Q-x_P,y_Q-y_P\rangle\)
\(\langle3,8\rangle=\langle4-x_P,1-y_P\rangle\)
\(3=4-x_P\)
\(x_P=1\)
\(8=1-y_P\)
\(y_P=-7\)
\(\text{The initial point is }(1,-7)\)
\(\textbf{8)}\) Find the initial point \((P)\) of vector \(\vec{v}\) from \(P\) to \(Q\) where \(\vec{v}= \langle-5,2\rangle \) and \( Q= (3,-1)\)
The initial point is \((8,-3)\)
\(\vec{PQ}=\langle x_Q-x_P,y_Q-y_P\rangle\)
\(\langle-5,2\rangle=\langle3-x_P,-1-y_P\rangle\)
\(-5=3-x_P\)
\(x_P=8\)
\(2=-1-y_P\)
\(y_P=-3\)
\(\text{The initial point is }(8,-3)\)
\(\textbf{9)}\) Find the initial point \((P)\) of vector \(\vec{v}\) from \(P\) to \(Q\) where \(\vec{v}= \langle-4,0\rangle \) and \( Q= (9,9)\)
The initial point is \((13,9)\)
\(\vec{PQ}=\langle x_Q-x_P,y_Q-y_P\rangle\)
\(\langle-4,0\rangle=\langle9-x_P,9-y_P\rangle\)
\(-4=9-x_P\)
\(x_P=13\)
\(0=9-y_P\)
\(y_P=9\)
\(\text{The initial point is }(13,9)\)
\(\textbf{10)}\) Find the displacement vector between \(A (-2,-3) \) and \( B (5,9)\)
\(\vec{AB}\) is \(\langle7,12\rangle\) or \(7\vec{i}+12\vec{j}\)
\(\vec{AB}=\langle x_2-x_1,y_2-y_1\rangle\)
\(\vec{AB}=\langle5-(-2),9-(-3)\rangle\)
\(\vec{AB}=\langle7,12\rangle\)
\(\vec{AB}=7\vec{i}+12\vec{j}\)
\(\textbf{11)}\) Find the displacement vector between \(A (10,-4) \) and \( B (-2,1)\)
\(\vec{AB}\) is \(\langle-12,5\rangle\) or \(-12\vec{i}+5\vec{j}\)
\(\vec{AB}=\langle x_2-x_1,y_2-y_1\rangle\)
\(\vec{AB}=\langle-2-10,1-(-4)\rangle\)
\(\vec{AB}=\langle-12,5\rangle\)
\(\vec{AB}=-12\vec{i}+5\vec{j}\)
\(\textbf{12)}\) Find the displacement vector between \(C (-1,4,7) \) and \( D (2,-6,10)\)
\(\vec{CD}\) is \(\langle3,-10,3\rangle\) or \(3\vec{i}-10\vec{j}+3\vec{k}\)
\(\vec{CD}=\langle x_2-x_1,y_2-y_1,z_2-z_1\rangle\)
\(\vec{CD}=\langle2-(-1),-6-4,10-7\rangle\)
\(\vec{CD}=\langle3,-10,3\rangle\)
\(\vec{CD}=3\vec{i}-10\vec{j}+3\vec{k}\)
\(\textbf{13)}\) Find the displacement vector between \(C (8,0,-5) \) and \( D (-1,6,2)\)
\(\vec{CD}\) is \(\langle-9,6,7\rangle\) or \(-9\vec{i}+6\vec{j}+7\vec{k}\)
\(\vec{CD}=\langle x_2-x_1,y_2-y_1,z_2-z_1\rangle\)
\(\vec{CD}=\langle-1-8,6-0,2-(-5)\rangle\)
\(\vec{CD}=\langle-9,6,7\rangle\)
\(\vec{CD}=-9\vec{i}+6\vec{j}+7\vec{k}\)
\(\textbf{14)}\) Find the terminal point \((Q)\) of vector \(\vec{v}\) from \(P\) to \(Q\) where \(\vec{v}= \langle4,-6\rangle \) and \( P= (-2,5)\)
The terminal point is \((2,-1)\)
\(\vec{PQ}=\langle x_Q-x_P,y_Q-y_P\rangle\)
\(\langle4,-6\rangle=\langle x_Q-(-2),y_Q-5\rangle\)
\(4=x_Q+2\)
\(x_Q=2\)
\(-6=y_Q-5\)
\(y_Q=-1\)
\(\text{The terminal point is }(2,-1)\)
\(\textbf{15)}\) Find the terminal point \((Q)\) of vector \(\vec{v}\) from \(P\) to \(Q\) where \(\vec{v}= \langle-3,7\rangle \) and \( P= (6,-2)\)
The terminal point is \((3,5)\)
\(\vec{PQ}=\langle x_Q-x_P,y_Q-y_P\rangle\)
\(\langle-3,7\rangle=\langle x_Q-6,y_Q-(-2)\rangle\)
\(-3=x_Q-6\)
\(x_Q=3\)
\(7=y_Q+2\)
\(y_Q=5\)
\(\text{The terminal point is }(3,5)\)
Challenge Problems
\(\textbf{16)}\) Find the initial point \((P)\) of vector \(\vec{v}\) from \(P\) to \(Q\) where \(\vec{v}= \langle2,-5,4\rangle \) and \( Q= (7,-1,10)\)
The initial point is \((5,4,6)\)
\(\vec{PQ}=\langle x_Q-x_P,y_Q-y_P,z_Q-z_P\rangle\)
\(\langle2,-5,4\rangle=\langle7-x_P,-1-y_P,10-z_P\rangle\)
\(2=7-x_P\)
\(x_P=5\)
\(-5=-1-y_P\)
\(y_P=4\)
\(4=10-z_P\)
\(z_P=6\)
\(\text{The initial point is }(5,4,6)\)
\(\textbf{17)}\) Find the terminal point \((Q)\) of vector \(\vec{v}\) from \(P\) to \(Q\) where \(\vec{v}= \langle-4,3,-8\rangle \) and \( P= (2,5,1)\)
The terminal point is \((-2,8,-7)\)
\(\vec{PQ}=\langle x_Q-x_P,y_Q-y_P,z_Q-z_P\rangle\)
\(\langle-4,3,-8\rangle=\langle x_Q-2,y_Q-5,z_Q-1\rangle\)
\(-4=x_Q-2\)
\(x_Q=-2\)
\(3=y_Q-5\)
\(y_Q=8\)
\(-8=z_Q-1\)
\(z_Q=-7\)
\(\text{The terminal point is }(-2,8,-7)\)
\(\textbf{18)}\) Find \(k\) if the displacement vector from \(A(2,k)\) to \(B(9,4)\) is \(\langle7,-3\rangle\).
The answer is \(k=7\)
\(\vec{AB}=\langle9-2,4-k\rangle\)
\(\vec{AB}=\langle7,4-k\rangle\)
\(\langle7,4-k\rangle=\langle7,-3\rangle\)
\(4-k=-3\)
\(-k=-7\)
\(k=7\)
\(\textbf{19)}\) Find \(k\) if the displacement vector from \(A(-1,5,2)\) to \(B(3,k,-4)\) is \(\langle4,-8,-6\rangle\).
The answer is \(k=-3\)
\(\vec{AB}=\langle3-(-1),k-5,-4-2\rangle\)
\(\vec{AB}=\langle4,k-5,-6\rangle\)
\(\langle4,k-5,-6\rangle=\langle4,-8,-6\rangle\)
\(k-5=-8\)
\(k=-3\)
\(\textbf{20)}\) Find the midpoint of the segment from \(A(-5,8)\) to \(B(6,0)\) using the displacement vector.
The midpoint is \(\left(\frac{1}{2},4\right)\)
\(\vec{AB}=\langle6-(-5),0-8\rangle\)
\(\vec{AB}=\langle11,-8\rangle\)
\(\text{Half of the displacement vector is }\left\langle\frac{11}{2},-4\right\rangle\)
\(\text{Add this to point }A.\)
\(\left(-5,8\right)+\left\langle\frac{11}{2},-4\right\rangle=\left(\frac{1}{2},4\right)\)
\(\text{The midpoint is }\left(\frac{1}{2},4\right)\)
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