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Question

Find a vector $\stackrel{\to }{a}$ of magnitude $5\sqrt{2}$, making an angle of $\frac{\mathrm{\pi }}{4}$ with x-axis, $\frac{\mathrm{\pi }}{2}$ with y-axis and an acute angle θ with z-axis. [CBSE 2014]

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Solution

It is given that vector $\stackrel{\to }{a}$ makes an angle of $\frac{\mathrm{\pi }}{4}$ with x-axis, $\frac{\mathrm{\pi }}{2}$ with y-axis and an acute angle θ with z-axis. $\therefore l=\mathrm{cos}\frac{\mathrm{\pi }}{4}=\frac{1}{\sqrt{2}},m=\mathrm{cos}\frac{\mathrm{\pi }}{2}=0,n=\mathrm{cos}\theta$ Now, ${l}^{2}+{m}^{2}+{n}^{2}=1\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}+0+{\mathrm{cos}}^{2}\theta =1\phantom{\rule{0ex}{0ex}}⇒{\mathrm{cos}}^{2}\theta =1-\frac{1}{2}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}\theta =\frac{1}{\sqrt{2}}\left(\theta \mathrm{is}\mathrm{acute}\right)$ We know that $\stackrel{\to }{a}=\left|\stackrel{\to }{a}\right|\left(l\stackrel{^}{i}+m\stackrel{^}{j}+n\stackrel{^}{k}\right)\phantom{\rule{0ex}{0ex}}⇒\stackrel{\to }{a}=5\sqrt{2}\left(\frac{1}{\sqrt{2}}\stackrel{^}{i}+0\stackrel{^}{j}+\frac{1}{\sqrt{2}}\stackrel{^}{k}\right)\phantom{\rule{0ex}{0ex}}⇒\stackrel{\to }{a}=5\left(\stackrel{^}{i}+0\stackrel{^}{j}+\stackrel{^}{k}\right)$

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