Solveeit Logo

Question

Mathematics Question on Vector Algebra

The sine of the angle between the vectors i2j+3k\overrightarrow{i}-2\overrightarrow{j}+3\overrightarrow{k} and 2i+j+k2\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k} is

A

527\frac{5}{2\sqrt7}

B

57\frac{5}{\sqrt7}

C

314\frac{3}{\sqrt14}

D

521\frac{5}{21}

Answer

527\frac{5}{2\sqrt7}

Explanation

Solution

cosθ=a.babcos\,\theta=\frac{\vec{a}.\vec{b}}{\left|\vec{a}\right|\left|\vec{b}\right|} =22+31+4+94+1+1=\frac{2-2+3}{\sqrt{1+4+9}\sqrt{4+1+1}} =384=328=\frac{3}{\sqrt{84}}=\frac{\sqrt{3}}{\sqrt{28}} sinθ=1cos2θ\therefore sin\,\theta=\sqrt{1-cos^{2}\,\theta} =1328=528=527=\sqrt{1-\frac{3}{28}}=\frac{5}{\sqrt{28}}=\frac{5}{2\sqrt{7}}