Question

A vector of magnitude 6, making angle $\frac{\mathrm{\pi }}{4}\mathrm{with}x-\mathrm{axis},\frac{\mathrm{\pi }}{3}$ with y-axis and an acute angle with z-axis is ____________.

Answer 1

Given: A vector of magnitude 6, making angle $\frac{\mathrm{\pi }}{4}\mathrm{with}x-\mathrm{axis},\frac{\mathrm{\pi }}{3}$ with y-axis and an acute angle with z-axis


Let a vector $\overrightarrow{a}$ makes an angle $\frac{\mathrm{\pi }}{4}\mathrm{with}x-\mathrm{axis},\frac{\mathrm{\pi }}{3}$ with y-axis and an acute angle with z-axis and is of magnitude 6.

$$\begin{gathered} \mathrm{Therefore}, \\ l=\cos \frac{\mathrm{\pi }}{4}=\frac{1}{\sqrt{2}} \\ m=\cos \frac{\mathrm{\pi }}{3}=\frac{1}{2} \\ n=\cos \theta \\ \mathrm{We}\mathrm{know}, \\ {l}^2+m^2+n^2=1 \\ \Rightarrow {\left(\frac{1}{\sqrt{2}}\right)}^2+{\left(\frac{1}{2}\right)}^2+{\left(\cos \theta \right)}^2=1 \\ \Rightarrow \frac{1}{2}+\frac{1}{4}+{\left(\cos \theta \right)}^2=1 \\ \Rightarrow {\left(\cos \theta \right)}^2=1-\frac{1}{2}-\frac{1}{4} \\ \Rightarrow {\left(\cos \theta \right)}^2=\frac{4-2-1}{4} \\ \Rightarrow {\left(\cos \theta \right)}^2=\frac{1}{4} \\ \Rightarrow \cos \theta =\frac{1}{2}\left(\because \theta \mathrm{is}\mathrm{acute}\right) \\ \Rightarrow n=\frac{1}{2} \\ Thus, \\ \overrightarrow{a}=\left|\overrightarrow{a}\right|\left(l\hat{i}+m\hat{j}+n\hat{k}\right) \\ =6\left(\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{2}\hat{j}+\frac{1}{2}\hat{k}\right) \\ =3\sqrt{2}\hat{i}+3\hat{j}+3\hat{k} \end{gathered}$$


Hence, a vector of magnitude 6, making angle $\frac{\mathrm{\pi }}{4}\mathrm{with}x-\mathrm{axis},\frac{\mathrm{\pi }}{3}$ with y−axis and an acute angle with z−axis is $\underline{3\sqrt{2}\hat{i}+3\hat{j}+3\hat{k}}.$