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A= i bar +2j bar +2k bar. Calculate angle between A bar and y axis.\(\vec A=\hat i+2\hat j+2\hat k\) |
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Answer» \(\vec A=\hat i+2\hat j+2\hat k\) Equation of y-axis is \(\vec y=0\hat i+\hat j + 0\hat k\) |\(\vec A\)| = \(\sqrt{1+4+4}=3\) |\(\vec y\)| = \(\sqrt{0+1+0}=1\) \(\vec A.\vec y=(\hat i+2\hat j+2\hat k).(0\hat i+\hat j+0\hat k)\) ⇒ |\(\vec A\)||\(\vec y\)| cos θ = 2, where θ is angle between \(\vec A\) and y - axis. ⇒ cos θ = \(\frac2{|\vec A||\vec y|}=\frac23\) ⇒ θ = cos-1(2/3) ≈ 48.2° |
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