If $\overline{\mathrm{a}}=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$ and $\overline{\mathrm{b}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}}$ are two vectors, then the angle between the vectors $3 \bar{a}+5 \bar{b}$ and $5 \bar{a}+3 \bar{b}$ is

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Aicharya · Accepted Answer

$\begin{aligned} 3 \overline{\mathrm{a}}+5 \overline{\mathrm{~b}} & =3(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+5(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}}) \ & =13 \hat{\mathrm{i}}+9 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}\end{aligned}$ $\begin{aligned} 5 \overline{\mathrm{a}}+3 \overline{\mathrm{~b}} & =5(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+3(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}}) \ & =11 \hat{\mathrm{i}}-\hat{\mathrm{j}}+12 \hat{\mathrm{k}}\end{aligned}$ $\begin{aligned} 	herefore \quad \cos 	heta & =\frac{(13 \hat{\mathrm{i}}+9 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \cdot(11 \hat{\mathrm{i}}-\hat{\mathrm{j}}+12 \hat{\mathrm{k}})}{\sqrt{169+81+16} \sqrt{121+1+144}} \ & =\frac{13(11)+9(-1)+1(12)}{\sqrt{266} \sqrt{266}}\end{aligned}$ $\begin{aligned} & =\frac{143-9+48}{266} \ & =\frac{182}{266} \ & =\frac{13}{19} \ 	herefore \quad 	heta & =\cos ^{-1}\left(\frac{13}{19}ight)\end{aligned}$

If $\overline{\mathrm{a}}=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$ and…

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