Vectors: demonstrate whether points are collinear

[K_Swiss]

Can you please check whether my answers are correct? If they aren't, how would you find if the points are parallel to one another with 3 coordinates?

Using vectors, demonstrate that these points are collinear.

a) P(15 , 10)
Q(6 , 4)
R(-12 , -8)

Vector PQ = (-9 , -6)
Vector QR = (-18 , -12)
Vector RP = (27 , -18)

(-9 / 27 / -18) = (-6 / -18 / -12)

Therefore, not collinear.

b)
D(33, -5, 20)
E(6, 4, -16)
F(9, 3, -12)

Vector DE = (-27, 9, -36)
Vector EF = (3, -1, 4)
Vector FD = (24, -8, 32)

(-27 / 3 / 24) = (9 / - 1 / -8) = (-36 / 4 / 32)

 

[soroban]

Hello, K_Swiss!

Two vectors are parallel if one is a multiple of the other.

Using vectors, demonstrate that these points are collinear.

\(\displaystyle a)\;P(15 , 10)\quad Q(6 , 4) \quad R(-12 , -8)\)

\(\displaystyle \overrightarrow{PQ} \;=\;\langle -9,-6\rangle \;=\;-3\,\langle3,2\rangle\)

\(\displaystyle \overrightarrow{QR} \;=\;\langle-18,-27\rangle \;=\;-6\,\langle3,2\rangle\)

\(\displaystyle \text{Hence: }\:\overrightarrow{PQ} \parallel \overrightarrow{QR}\)


\(\displaystyle \text{Since }\overrightarrow{PQ}\text{ and }\overrightarrow{QR}\text{ share point }Q\!:\;P,Q,R\text{ are collinear.}\)

\(\displaystyle b)\;\;D(33, -5, 20)\quad E(6, 4, -16) \quad F(9, 3, -12)\)

\(\displaystyle \overrightarrow{DE} \;=\;\langle -27,9,-36\rangle \;=\;-9\,\langle 3,-1,4\rangle\)

\(\displaystyle \overrightarrow{EF} \;=\;\langle 3,-1,4\rangle\)

\(\displaystyle \text{Hence: }\:\overrightarrow{DE} \parallel \overrightarrow{EF}\)

\(\displaystyle \text{Since }\overrightarrow{DE}\text{ and }\overrightarrow{EF}\text{ share point }E\!:\;D,E,F\text{ are collinear.}\)