Parallel vectors point in the same or opposite direction, so one vector is a scalar multiple of the other. Perpendicular vectors meet at a right angle, so their dot product is equal to \(0\). These problems include identifying parallel, perpendicular, or neither relationships and solving for missing values that create those relationships.
Notes
Parallel Vectors
\(\langle x_1,y_1 \rangle \text{ and } \langle x_2,y_2 \rangle \text{ are parallel if}\)
\(\langle x_2,y_2 \rangle = \langle kx_1,ky_1 \rangle\)
Perpendicular Vectors
\(\langle x_1,y_1 \rangle \text{ and } \langle x_2,y_2 \rangle \text{ are perpendicular if}\)
\(x_1 x_2+ y_1 y_2 =0\)
Practice Questions
Are the following pairs of vectors parallel, perpendicular or neither?
\(\textbf{1)}\) \(\langle8,2 \rangle \text{ and } \langle-4,-1 \rangle\)
They are parallel
\(\langle-4,-1\rangle=-\frac{1}{2}\langle8,2\rangle\)
\(\text{Since one vector is a scalar multiple of the other, the vectors are parallel.}\)
\(\text{They are parallel.}\)
\(\textbf{2)}\) \(\langle3,6 \rangle \text{ and } \langle1,-2 \rangle\)
They are neither
\(\frac{1}{3}\neq \frac{-2}{6}\)
\(\text{So the vectors are not parallel.}\)
\(\langle3,6\rangle\cdot\langle1,-2\rangle=(3)(1)+(6)(-2)\)
\(3-12=-9\)
\(\text{Since the dot product is not }0,\text{ the vectors are not perpendicular.}\)
\(\text{They are neither.}\)
\(\textbf{3)}\) \(\langle4,-8 \rangle \text{ and } \langle4,2 \rangle\)
They are perpendicular
\(\langle4,-8\rangle\cdot\langle4,2\rangle=(4)(4)+(-8)(2)\)
\(16-16=0\)
\(\text{Since the dot product is }0,\text{ the vectors are perpendicular.}\)
\(\text{They are perpendicular.}\)
\(\textbf{4)}\) \(\langle1,2 \rangle \text{ and } \langle3,4 \rangle\)
They are neither
\(\frac{3}{1}\neq \frac{4}{2}\)
\(\text{So the vectors are not parallel.}\)
\(\langle1,2\rangle\cdot\langle3,4\rangle=(1)(3)+(2)(4)\)
\(3+8=11\)
\(\text{Since the dot product is not }0,\text{ the vectors are not perpendicular.}\)
\(\text{They are neither.}\)
\(\textbf{5)}\) \(\langle-2,4 \rangle \text{ and } \langle4,-8 \rangle\)
They are parallel
\(\langle4,-8\rangle=-2\langle-2,4\rangle\)
\(\text{Since one vector is a scalar multiple of the other, the vectors are parallel.}\)
\(\text{They are parallel.}\)
\(\textbf{6)}\) \(\langle3,1 \rangle \text{ and } \langle9,3 \rangle\)
They are parallel
\(\langle9,3\rangle=3\langle3,1\rangle\)
\(\text{Since one vector is a scalar multiple of the other, the vectors are parallel.}\)
\(\text{They are parallel.}\)
\(\textbf{7)}\) \(\langle6,-2 \rangle \text{ and } \langle1,3 \rangle\)
They are perpendicular
\(\langle6,-2\rangle\cdot\langle1,3\rangle=(6)(1)+(-2)(3)\)
\(6-6=0\)
\(\text{Since the dot product is }0,\text{ the vectors are perpendicular.}\)
\(\text{They are perpendicular.}\)
\(\textbf{8)}\) \(\langle5,10 \rangle \text{ and } \langle1,2 \rangle\)
They are parallel
\(\langle5,10\rangle=5\langle1,2\rangle\)
\(\text{Since one vector is a scalar multiple of the other, the vectors are parallel.}\)
\(\text{They are parallel.}\)
\(\textbf{9)}\) \(\langle2,7 \rangle \text{ and } \langle7,-2 \rangle\)
They are perpendicular
\(\langle2,7\rangle\cdot\langle7,-2\rangle=(2)(7)+(7)(-2)\)
\(14-14=0\)
\(\text{Since the dot product is }0,\text{ the vectors are perpendicular.}\)
\(\text{They are perpendicular.}\)
\(\textbf{10)}\) \(\langle-6,9 \rangle \text{ and } \langle2,-3 \rangle\)
They are parallel
\(\langle-6,9\rangle=-3\langle2,-3\rangle\)
\(\text{Since one vector is a scalar multiple of the other, the vectors are parallel.}\)
\(\text{They are parallel.}\)
\(\textbf{11)}\) \(\langle4,1 \rangle \text{ and } \langle-2,8 \rangle\)
They are perpendicular
\(\langle4,1\rangle\cdot\langle-2,8\rangle=(4)(-2)+(1)(8)\)
\(-8+8=0\)
\(\text{Since the dot product is }0,\text{ the vectors are perpendicular.}\)
\(\text{They are perpendicular.}\)
\(\textbf{12)}\) \(\langle3,-5 \rangle \text{ and } \langle6,-10 \rangle\)
They are parallel
\(\langle6,-10\rangle=2\langle3,-5\rangle\)
\(\text{Since one vector is a scalar multiple of the other, the vectors are parallel.}\)
\(\text{They are parallel.}\)
\(\textbf{13)}\) \(\langle1,4 \rangle \text{ and } \langle2,5 \rangle\)
They are neither
\(\frac{2}{1}\neq\frac{5}{4}\)
\(\text{So the vectors are not parallel.}\)
\(\langle1,4\rangle\cdot\langle2,5\rangle=(1)(2)+(4)(5)\)
\(2+20=22\)
\(\text{Since the dot product is not }0,\text{ the vectors are not perpendicular.}\)
\(\text{They are neither.}\)
\(\textbf{14)}\) \(\langle0,5 \rangle \text{ and } \langle3,0 \rangle\)
They are perpendicular
\(\langle0,5\rangle\cdot\langle3,0\rangle=(0)(3)+(5)(0)\)
\(0+0=0\)
\(\text{Since the dot product is }0,\text{ the vectors are perpendicular.}\)
\(\text{They are perpendicular.}\)
\(\textbf{15)}\) \(\langle-1,6 \rangle \text{ and } \langle3,2 \rangle\)
They are neither
\(\frac{3}{-1}\neq\frac{2}{6}\)
\(\text{So the vectors are not parallel.}\)
\(\langle-1,6\rangle\cdot\langle3,2\rangle=(-1)(3)+(6)(2)\)
\(-3+12=9\)
\(\text{Since the dot product is not }0,\text{ the vectors are not perpendicular.}\)
\(\text{They are neither.}\)
Challenge Problems
\(\textbf{16)}\) Find \(k\) so that \(\langle k,6\rangle\) and \(\langle2,3\rangle\) are parallel.
The answer is \(k=4\)
\(\text{If the vectors are parallel, then one is a scalar multiple of the other.}\)
\(\langle k,6\rangle=c\langle2,3\rangle\)
\(6=3c\)
\(c=2\)
\(k=2c\)
\(k=4\)
\(\text{The answer is }k=4\)
\(\textbf{17)}\) Find \(k\) so that \(\langle4,k\rangle\) and \(\langle3,6\rangle\) are perpendicular.
The answer is \(k=-2\)
\(\text{If the vectors are perpendicular, then their dot product is }0.\)
\(\langle4,k\rangle\cdot\langle3,6\rangle=0\)
\((4)(3)+(k)(6)=0\)
\(12+6k=0\)
\(6k=-12\)
\(k=-2\)
\(\text{The answer is }k=-2\)
\(\textbf{18)}\) Find \(k\) so that \(\langle5,-10\rangle\) and \(\langle k,-2\rangle\) are parallel.
The answer is \(k=1\)
\(\text{If the vectors are parallel, then one is a scalar multiple of the other.}\)
\(\langle5,-10\rangle=c\langle k,-2\rangle\)
\(-10=c(-2)\)
\(c=5\)
\(5=5k\)
\(k=1\)
\(\text{The answer is }k=1\)
\(\textbf{19)}\) Find \(k\) so that \(\langle k,8\rangle\) and \(\langle4,-3\rangle\) are perpendicular.
The answer is \(k=6\)
\(\text{If the vectors are perpendicular, then their dot product is }0.\)
\(\langle k,8\rangle\cdot\langle4,-3\rangle=0\)
\((k)(4)+(8)(-3)=0\)
\(4k-24=0\)
\(4k=24\)
\(k=6\)
\(\text{The answer is }k=6\)
\(\textbf{20)}\) Find \(k\) so that \(\langle2,k\rangle\) and \(\langle6,15\rangle\) are parallel.
The answer is \(k=5\)
\(\text{If the vectors are parallel, then one is a scalar multiple of the other.}\)
\(\langle6,15\rangle=c\langle2,k\rangle\)
\(6=2c\)
\(c=3\)
\(15=3k\)
\(k=5\)
\(\text{The answer is }k=5\)
See Related Pages\(\)
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