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Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
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If \(\overrightarrow a = \hat i + \hat j + \hat k, \overrightarrow b = 2\hat i - \hat j + 3\hat k\) and \(\overrightarrow c = \hat i - 2\hat j + \hat k\), find a unit vector parallel to the vector \(2\overrightarrow a - \overrightarrow b + 3\overrightarrow c\)

$\begin{array}{1 1} (A) \large\frac{1}{\sqrt{22}}(3\hat i +3\hat j+2\hat k) \\ (B) \large\frac{1}{\sqrt{22}}(3\hat i -3\hat j-2\hat k) \\ (C) \large\frac{1}{\sqrt{22}}(3\hat i +3\hat j-2\hat k) \\ (D) \large\frac{1}{\sqrt{22}}(3\hat i -3\hat j+2\hat k) \end{array} $

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1 Answer

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Toolbox:
  • Unit vector of $\overrightarrow a=\large\frac{\overrightarrow a}{|\overrightarrow a|}$
  • $|\overrightarrow a|=\sqrt{a_1^2+a_2^2+a_3^2}$ where $\overrightarrow a=a_1\hat i+a_2\hat j+a_3\hat k$
Step 1:
$\overrightarrow a=\hat i+\hat j+\hat k,\overrightarrow b=2\hat i-\hat j+3\hat k$ and $\overrightarrow c=\hat i-2\hat j+\hat k$
$2\overrightarrow a=2(\hat i+\hat j+\hat k)$
$\quad\;\;=2\hat i+2\hat j+2\hat k$
$3\overrightarrow c=3(\hat i-2\hat j+\hat k)$
$\quad\;\;=3\hat i-6\hat j+3\hat k$
Step 2:
We have to find $2\overrightarrow a-\overrightarrow b+3\overrightarrow c$
$2\overrightarrow a-\overrightarrow b+3\overrightarrow c=(2\hat i+2\hat j+2\hat k)-(2\hat i-\hat j+3\hat k)+(3\hat i-6\hat j+3\hat k)$
$\qquad\qquad\qquad=3\hat i-3\hat j+2\hat k$
Step 3:
Next we have to find the magnitude of this vector $|2\overrightarrow a-\overrightarrow b+3\overrightarrow c$|
$|2\overrightarrow a-\overrightarrow b+3\overrightarrow c|=\sqrt{(3)^2+(-3)^2+(2)^2}$
$\qquad\qquad\quad\;\;\;\;=\sqrt{9+9+4}$
$\qquad\qquad\quad\;\;\;\;=\sqrt{22}$
Step 4:
Hence the unit vector parallel to $2\overrightarrow a-\overrightarrow b+3\overrightarrow c=\large\frac{2\overrightarrow a-\overrightarrow b+3\overrightarrow c}{|2\overrightarrow a-\overrightarrow b+3\overrightarrow c|}$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad=\large\frac{1}{\sqrt{22}}$$(3\hat i-3\hat j+2\hat k)$
answered May 23, 2013 by sreemathi.v
 

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