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Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
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Find a vector of magnitude $5$ units, and parallel to the resultant of the vectors \( \overrightarrow a = 2\hat i + 3 \hat j - \hat k\) and \(\overrightarrow b = \hat i - 2\hat j + \hat k\)

$\begin{array}{1 1} (A) \large\frac{3\sqrt {10}}{2}\hat i-\large\frac{3\sqrt{10}}{2}\hat j \\ (B) \large\frac{3\sqrt {10}}{2}\hat i+\large\frac{3\sqrt{10}}{2}\hat j \\ (C) \large\frac{-3\sqrt {10}}{2}\hat i-\large\frac{3\sqrt{10}}{2}\hat j \\ (D) \large\frac{-3\sqrt {10}}{2}\hat i+\large\frac{3\sqrt{10}}{2}\hat j \end{array} $

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  • Resultant of two vectors $\overrightarrow a\:and \: \overrightarrow b$ is given by $\overrightarrow c =\overrightarrow a + \overrightarrow b$
  • $|\overrightarrow a|=\sqrt{a_1^2+a_2^2+a_3^2}$ where $\overrightarrow a=a_1\hat i+a_2\hat j+a_3\hat k$
Step 1:
Let $\overrightarrow c$ be the resultant of $\overrightarrow a$ and $\overrightarrow b$.
Given :$\overrightarrow a=2\hat i+3\hat j-\hat k$ and $\overrightarrow b=\hat i-2\hat j+\hat k$
$\overrightarrow c =\overrightarrow a + \overrightarrow b$
$\quad=(2\hat i+3\hat j-\hat k)+(\hat i-2\hat j+\hat k)$
$\quad=3\hat i+\hat j$
Hence its magnitude $|\overrightarrow c|=\sqrt{3^2+1^2}$
Step 2:
We know unit vector is $\hat n=\large\frac{\overrightarrow a}{|\overrightarrow a|}$
Therefore unit vector of $\overrightarrow c$ = $\large\frac{1}{\sqrt{10}}$$(3\hat i+\hat j)$
$\qquad\qquad\qquad\qquad\quad\;=\large\frac{3}{\sqrt{10}}$$\hat i+\large\frac{3}{\sqrt{10}}$$\hat j$
Step 3:
But it is given $\overrightarrow c$ has a magnitude of 5.
Therefore $5[\big(\large\frac{3}{\sqrt{10}}$$\hat i\big)+\big(\large\frac{3}{\sqrt{10}}$$\hat j\big)]=\large\frac{15}{\sqrt{10}}$$\hat i+\large\frac{15}{\sqrt{10}}$$\hat j$
Step 4:
Rationalizing the denominator we get
$\quad=\large\frac{15\sqrt{10}}{10}$$\hat i+\large\frac{15\sqrt{10}}{10}$$\hat j$
$\quad=\large\frac{3\sqrt{10}}{2}$$\hat i+\large\frac{3\sqrt{10}}{2}$$\hat j$
Hence the required vector $\overrightarrow c=\large\frac{3\sqrt{10}}{2}$$\hat i+\large\frac{3\sqrt{10}}{2}$$\hat j$
answered May 23, 2013 by sreemathi.v

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