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  • Find the sine of the angle between the vectors a =3i +j +2k and b =2i minus 2j +4k

Find the sine of the angle between the vectors \overrightarrow{\mathrm{a}}=3 \hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}_{\text {and }} \overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}

 

Answers (1)

We have

\overrightarrow{\mathrm{a}}=3 \hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}_{\text {and }} \overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}

Let these vectors be represented as,

\begin{array}{l} \overrightarrow{\mathrm{a}}=\mathrm{a}_{1} \hat{\imath}+\mathrm{a}_{2} \hat{\jmath}+\mathrm{a}_{3} \hat{\mathrm{k}} \\ \overrightarrow{\mathrm{b}}=\mathrm{b}_{1} \hat{\imath}+\mathrm{b}_{2} \hat{\jmath}+\mathrm{b}_{3} \hat{\mathrm{k}} \end{array}

Comparing the vectors, we get

\begin{array}{l} a_{1}=3, a_{2}=1 \text { and } a_{3}=2 \\ b_{1}=2, b_{2}=-2 \text { and } b_{3}=4 \end{array}

To find sine of the angle between the vectors\vec{a} \text { and } \vec{b}, we can first find out cosine of the angle between them.

Cosine of the angle between \vec{a} \text { and } \vec{b}   is given by,

\\\cos \theta=\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{a}}| \overrightarrow{\mathrm{b}} \mid} \\ =\frac{(3 \hat{\imath}+\hat{\jmath}+2 \hat{\mathrm{k}})(2 \hat{\imath}-2 \hat{\jmath}+4 \hat{\mathrm{k}})}{|3 \hat{\imath}+\hat{\jmath}+2 \hat{\mathrm{k}}||2 \hat{\imath}-2 \hat{\jmath}+4 \hat{\mathrm{k}}|} \\ =\frac{(3 \hat{\imath})(2 \hat{\imath})+(\hat{j})(-2 \hat{\jmath})+(2 \hat{\mathrm{k}})(4 \hat{\mathrm{k}})}{\sqrt{3^{2}+1^{2}+2^{2}} \sqrt{2^{2}+(-2)^{2}+4^{2}}}

\\ \begin{aligned} &\because\text {we know that, } \hat{i} \times \hat{i}=\hat{\jmath} \times \hat{\jmath}=\hat{\mathrm{k}} \times \hat{\mathrm{k}}=1_{\text {and }} \hat{i} \times \hat{\mathrm{j}}=\hat{i} \times \hat{\mathrm{k}}=\hat{\mathrm{j}} \times \hat{\mathrm{k}}=0\\ &\text { So, }\left(a_{1} \hat{\imath}+a_{2} \hat{\jmath}+a_{3} \hat{k}\right)\left(b_{1} \hat{\imath}+b_{2} \hat{\jmath}+b_{3} \hat{k}\right)=a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}\\ &\text { Also, we know that, } \left.|a \hat{\imath}+b \hat{\jmath}+c \hat{k}|=\sqrt{a^{2}+b^{2}+c^{2}}\right] \end{aligned}

\\ =\frac{6-2+8}{\sqrt{9+1+4} \sqrt{4+4+16}} \\ =\frac{12}{\sqrt{14} \sqrt{24}} \\ =\frac{12}{2 \sqrt{14} \sqrt{6}} \\ =\frac{6}{\sqrt{14 \times 6}}

\\ =\frac{6}{\sqrt{84}} \\ =\frac{6}{2 \sqrt{21}} \\ =\frac{3}{\sqrt{21}}

By algebraic identity, we have

\\ \sin \theta=\sqrt{1-\cos ^{2} \theta} \\ =\sqrt{1-\left(\frac{3}{\sqrt{21}}\right)^{2}} \\ =\sqrt{1-\frac{9}{21}} \\ =\sqrt{\frac{21-9}{21}}

\\ =\sqrt{\frac{12}{21}} \\ =\sqrt{\frac{4}{7}} \\ =\frac{2}{\sqrt{7}}

Thus, sine of the angle between the vectors is \frac{2}{\sqrt{7}} .

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