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Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
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If \( \overrightarrow a, \overrightarrow b, \overrightarrow c, \) are mutually perpendicular vectors of equal magnitudes, show that the vector \( \overrightarrow a + \overrightarrow b + \overrightarrow c\) is equally inclined to \( \overrightarrow a, \overrightarrow b and \overrightarrow c\).

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1 Answer

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  • $\overrightarrow a.\overrightarrow =|\overrightarrow a|.\overrightarrow b|\cos\theta$
  • $\cos\theta=\large\frac{\overrightarrow a.\overrightarrow b}{|\overrightarrow a|\overrightarrow b|}$
Step 1:
It is given that $|\overrightarrow a|=|\overrightarrow b|=|\overrightarrow c|$
Let us assume it to be equal to some value $\lambda$
Therefore $\overrightarrow a|=|\overrightarrow b|=|\overrightarrow c|=\lambda$
Also $\overrightarrow a,\overrightarrow b$ and $\overrightarrow c$ are mutually perpendicular to each other.
Therefore $\overrightarrow a.\overrightarrow b=\overrightarrow b.\overrightarrow c=\overrightarrow c.\overrightarrow a=0$
Step 2:
Now let us take $\theta$ to be the angle between $\overrightarrow a$ and $\overrightarrow a+\overrightarrow b+\overrightarrow c$
Hence $\overrightarrow a.(\overrightarrow a+\overrightarrow b+\overrightarrow c)=|\overrightarrow a||\overrightarrow a+\overrightarrow b+\overrightarrow c|\cos\theta$
But $|\overrightarrow a|=\lambda$
Therefore $\overrightarrow a.(\overrightarrow a+\overrightarrow b+\overrightarrow c)=\lambda|\overrightarrow a+\overrightarrow b+\overrightarrow c|\cos\theta$--------(1)
Step 3:
Also $\overrightarrow a.(\overrightarrow a+\overrightarrow b+\overrightarrow c)=\overrightarrow a.\overrightarrow a+\overrightarrow a.\overrightarrow b+\overrightarrow a.\overrightarrow c$
$\overrightarrow a.\overrightarrow a=|\overrightarrow a|^2$ and $\overrightarrow a.\overrightarrow b=\overrightarrow a.\overrightarrow c=0$
Therefore $\overrightarrow a.(\overrightarrow a+\overrightarrow b+\overrightarrow c)=|\overrightarrow a|^2+0+0$
$\overrightarrow a.(\overrightarrow a+\overrightarrow b+\overrightarrow c)=\lambda^2$-----(2)
Step 4:
Equating (1) and (2) we get
$\lambda^2=\lambda|\overrightarrow a+\overrightarrow b+\overrightarrow c|\cos\theta$
$\cos\theta=\large\frac{\lambda}{|\overrightarrow a+\overrightarrow b+\overrightarrow c|}$
$\Rightarrow \theta=\cos^{-1}\bigg(\large\frac{\lambda}{|\overrightarrow a+\overrightarrow b+\overrightarrow c|}\bigg)$
Step 5:
Similarly we can prove that angle $\theta$ between $\overrightarrow b$ and $(\overrightarrow a+\overrightarrow b+\overrightarrow c)=\cos^{-1}\bigg(\large\frac{\lambda}{|\overrightarrow a+\overrightarrow b+\overrightarrow c|}\bigg)$
Angle $\theta$ between $\overrightarrow c$ and $(\overrightarrow a+\overrightarrow b+\overrightarrow c)=\cos^{-1}\bigg(\large\frac{\lambda}{|\overrightarrow a+\overrightarrow b+\overrightarrow c|}\bigg)$
Hence $(\overrightarrow a+\overrightarrow b+\overrightarrow c)$ is equally inclined with $\overrightarrow a,\overrightarrow b$ and $\overrightarrow c$.
answered May 27, 2013 by sreemathi.v
 

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