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Home  >>  TN XII Math  >>  Vector Algebra
+1 vote

If $\overrightarrow{a}=\overrightarrow{i}+\overrightarrow{3j}-\overrightarrow{2k}$ and $\overrightarrow{b}=\overrightarrow{-i}+\overrightarrow{3k}$ than find $\overrightarrow{a}\times\overrightarrow{b}$. Verify that $\overrightarrow{a}$ and $\overrightarrow{b}$ are perpendicular to$\overrightarrow{a}\times\overrightarrow{b}$ separately.

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  • $ \overrightarrow a.\overrightarrow b=| \overrightarrow a|| \overrightarrow b| \cos \theta$ $ \therefore \cos \theta = \large\frac{ \overrightarrow a. \overrightarrow b}{| \overrightarrow a|| \overrightarrow b|} \Rightarrow \theta = \cos^{-1} \large\frac{ \overrightarrow a. \overrightarrow b}{| \overrightarrow a|| \overrightarrow b|}$
  • If $ \overrightarrow a = a_1\overrightarrow i+a_2\overrightarrow j+a_3\overrightarrow k, \: \overrightarrow b = b_1\overrightarrow i+b_2\overrightarrow j+b_3\overrightarrow k$ then $ \overrightarrow a$ x $ \overrightarrow b = \begin{vmatrix} \overrightarrow i & \overrightarrow j & \overrightarrow k \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}$
Step 1
$ \overrightarrow a = \overrightarrow i+3\overrightarrow j-2\overrightarrow k\: \: \: \overrightarrow b = -\overrightarrow i+3\overrightarrow k$
$ \overrightarrow a \times \overrightarrow b = \begin{vmatrix} \overrightarrow i & \overrightarrow j & \overrightarrow k \\ 1 & 3 & -2 \\ -1 & 0 & 3 \end{vmatrix} = (9-0)\overrightarrow i-(3-2)\overrightarrow j+(0+3)\overrightarrow k$
$ = 9\overrightarrow i-\overrightarrow j+3\overrightarrow k$
Step 2
$ \overrightarrow a.(\overrightarrow a \times \overrightarrow b) =(\overrightarrow i+3\overrightarrow j-2\overrightarrow k).(9\overrightarrow i-\overrightarrow j+3\overrightarrow k)$
$ = (1)(9)+(3)(-1)+(-2)(3)$
$=9-3-6=0$
$ \overrightarrow b.(\overrightarrow a \times \overrightarrow b) = (-\overrightarrow i+3\overrightarrow k).(9\overrightarrow i-\overrightarrow j+3\overrightarrow k)$
$ = (-1)(9)+(3)(3)=-9+9=0$
Since $ \overrightarrow a.( \overrightarrow a \times \overrightarrow b)=0\: and \: \overrightarrow b.(\overrightarrow a \times \overrightarrow b)=0$ it is verified that $ \overrightarrow a\: and \: \overrightarrow a \times \overrightarrow b, \: and \: \overrightarrow b \: and \: \overrightarrow a \times \overrightarrow b\: are \perp$
answered Jun 5, 2013 by thanvigandhi_1
 

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