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Vectors: magnitude, direction and products

Practice section 3 of 9

Questions 13–18 in this collection

Practice questions

Worked example · 2 of 6

Combine a known angle and a right-angle condition

The angle between a\mathbf a and b\mathbf b is π/3\pi/3 radians, and the angle between b\mathbf b and a−b\mathbf a-\mathbf b is a right angle. Show that ∣a∣=2∣b∣|\mathbf a|=2|\mathbf b|.

2003/NYJC/I/2

CJC · Vectors Basic · Example 11

Solution

Self-practice · 4 of 6

Calculate two angles in degrees and radians

Calculate the angle between a\mathbf a and b\mathbf b if (i) a=2i−3j+6k\mathbf a=2\mathbf i-3\mathbf j+6\mathbf k, b=i+j\mathbf b=\mathbf i+\mathbf j; (ii) a=2i−4j+5k\mathbf a=2\mathbf i-4\mathbf j+5\mathbf k, b=i+3j+8k\mathbf b=\mathbf i+3\mathbf j+8\mathbf k.

CJC · Vectors Basic · Self-Practice 3

Solution

Worked example · 5 of 6

Choose both angle vectors from the same vertex

A,B,CA,B,C have position vectors i+2j+3k\mathbf i+2\mathbf j+3\mathbf k, i+j+k\mathbf i+\mathbf j+\mathbf k and 2i+3j−2k2\mathbf i+3\mathbf j-2\mathbf k respectively. Find ∠ABC\angle ABC to the nearest 0.1∘0.1^\circ.

CJC · Vectors Basic · Example 13

Solution

Self-practice · 6 of 6

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

Solution

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