Porisma

Worked solution

Check which scalar-product angle identities are valid

Which of the following is/are correct?

(1) PQ→⋅QR→=∣PQ→∣∣QR→∣cos⁡∠PQR\overrightarrow{PQ}\cdot\overrightarrow{QR}=|\overrightarrow{PQ}||\overrightarrow{QR}|\cos\angle PQR.

(2) BA→⋅BC→=∣BA→∣∣BC→∣cos⁡∠ABC\overrightarrow{BA}\cdot\overrightarrow{BC}=|\overrightarrow{BA}||\overrightarrow{BC}|\cos\angle ABC.

(3) KM→⋅NM→=∣KM→∣∣NM→∣cos⁡∠KMN\overrightarrow{KM}\cdot\overrightarrow{NM}=|\overrightarrow{KM}||\overrightarrow{NM}|\cos\angle KMN.

CJC · Vectors Basic · Self-Practice 4

Answer

(2) and (3) only.

Compare your reasoning

Check (1) at Q

PQ→⋅QR→=−∣PQ→∣∣QR→∣cos⁡∠PQR\begin{aligned}&\overrightarrow{PQ}\cdot\overrightarrow{QR}\\ & =-|\overrightarrow{PQ}||\overrightarrow{QR}|\cos\angle PQR\end{aligned}

PQ points into Q while QR points out. This is not an identity for the stated angle. Use QP and QR, or reverse both of those correct vectors.

Check (2) at B

BA→⋅BC→=∣BA→∣∣BC→∣cos⁡∠ABC\begin{aligned}&\overrightarrow{BA}\cdot\overrightarrow{BC}\\ & =|\overrightarrow{BA}||\overrightarrow{BC}|\cos\angle ABC\end{aligned}

BA and BC both point away from B, so the displayed formula is correct.

Check (3) at M

KM→⋅NM→=(−MK→)⋅(−MN→)=MK→⋅MN→\begin{aligned}&\overrightarrow{KM}\cdot\overrightarrow{NM}\\ & =(-\overrightarrow{MK})\cdot(-\overrightarrow{MN})\\ & =\overrightarrow{MK}\cdot\overrightarrow{MN}\end{aligned}

KM and NM both point towards M. Reversing both leaves the dot product and angle unchanged.

How did this question go?

Solution notes

The printed selection (2) and (3) only is retained; these explanations are authored. Assume nonzero segments so each named angle is defined. Formula (1) can coincide numerically at a right angle, but it is not a generally correct identity. Student arrow notation is retained; the Teacher rendering of (1) omits those arrow marks.