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

Find the vectors whose length $5$ and which are perpendicular to the vectors $\overrightarrow{a}=\overrightarrow{3i}+\overrightarrow{j}-\overrightarrow{4k}$ and $\overrightarrow{b}=\overrightarrow{6i}+\overrightarrow{5j}-\overrightarrow{2k}$

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  • For two vectors $ \overrightarrow a \: and \: \overrightarrow b$, the vector product $ \overrightarrow a$ x $ \overrightarrow b=|\overrightarrow a||\overrightarrow b| \sin \theta \hat n$ with $ \hat n \perp $ to $ \overrightarrow a \: and \: \overrightarrow b\: and \: \overrightarrow a, \overrightarrow b, \hat n$ forming a right handed system.
  • 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
The required vectors are $ \pm 5 \large\frac{\overrightarrow a \times \overrightarrow b}{|\overrightarrow a \times \overrightarrow b|}$
Step 2
$ \overrightarrow a \times \overrightarrow b = \begin{vmatrix} \overrightarrow i & \overrightarrow j & \overrightarrow k \\ 3 & 1 & -4 \\ 6 & 5 & -2 \end{vmatrix} = (-2+20)\overrightarrow i-(-6+24)\overrightarrow j+(15-6)\overrightarrow k$
$ = -18\overrightarrow i-18\overrightarrow j+9\overrightarrow k$
$ = -9(2\overrightarrow i+2\overrightarrow j-\overrightarrow k)$
Step 3
$ | \overrightarrow a \times \overrightarrow b|=9 \sqrt{4+4+1}=27$
The required vectors are $ \pm \large\frac{5(-9)(2\overrightarrow i+2\overrightarrow j-\overrightarrow k)}{27}$
$ = \pm \large\frac{5}{3}(2\overrightarrow i+2\overrightarrow j-\overrightarrow k)$

 

answered Jun 5, 2013 by thanvigandhi_1
edited Jun 23, 2013 by thanvigandhi_1
 

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