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Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
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Find the angle between the vectors \(\hat i - 2\hat j + 3\hat k\) and \( 3\hat i - 2\hat j + \hat k\)

$\begin{array}{1 1} (A) \cos^{-1} \bigg( \large\frac{2}{7} \bigg) \\ (B) \cos^{-1} \bigg( \large\frac{4}{7} \bigg) \\ (C) \cos^{-1} \bigg( \large\frac{1}{7} \bigg) \\ (D) \cos^{-1} \bigg( \large\frac{5}{7} \bigg) \end{array} $

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1 Answer

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Toolbox:
  • The scalar product of two vectors is $\overrightarrow a.\overrightarrow b=\mid \overrightarrow a\mid\mid\overrightarrow b\mid\cos\theta$
  • Hence the angle between two vectors is given by $\cos\theta=\large\frac{\overrightarrow a.\overrightarrow b}{\mid\overrightarrow a\mid\mid\overrightarrow b\mid}$
Step 1:
Let $\overrightarrow a=\hat i-2\hat j+3\hat k$ and $\overrightarrow b=3\hat i-2\hat j+\hat k$
We know that $\overrightarrow a.\overrightarrow b=|\overrightarrow a||\overrightarrow b|\cos\theta$
$\cos\theta=\large\frac{\overrightarrow a.\overrightarrow b}{\mid\overrightarrow a\mid\mid\overrightarrow b\mid}$
$\mid \overrightarrow a\mid=\sqrt{1^2+(-2)^2+3^2}$
$\qquad=\sqrt{1+4+9}$
$\qquad=\sqrt{14}$
Similarly $\mid \overrightarrow b\mid=\sqrt{3^2+(-2)^2+1^2}$
$\qquad=\sqrt{9+4+1}$
$\qquad=\sqrt{14}$
Step 2:
$\overrightarrow a.\overrightarrow b=(\hat i-2\hat j+3\hat k).(3\hat i-2\hat j+\hat k)$
Scalar product of two vectors is just multiplication of the scalar quantities of the respective components.
$\overrightarrow a.\overrightarrow b=(3+4+3)=10.$
Step 3:
Substituting these values for $\cos\theta$ we get,
$\cos\theta=\large\frac{10}{\sqrt{14}\sqrt {14}}$
$\qquad=\large\frac{10}{14}$
$\qquad=\large\frac{5}{7}$
Therefore $\theta=\cos^{-1}\big(\large\frac{5}{7}\big)$
answered May 20, 2013 by sreemathi.v
 

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