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Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
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Find the unit vector in the direction of vector \( \overline{PQ}\), where $P$ and $Q$ are the points $(1, 2, 3)$ and $(4, 5, 6)$ respectively.

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

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  • If the initial point A is $(x_1,y_1,z_1)$ and terminal point is $(x_2,y_2,z_2)$
  • Then $\overrightarrow{AB}=(x_2-x_1)\hat i+(y_2-y_1)\hat j+(z_2-z_1)\hat k$
  • Unit vector in the direction of $\overrightarrow{a}$ is given by $\hat a=\large\frac{\overrightarrow a}{|\overrightarrow a|}$
Step 1:
Let $\overrightarrow{OP}=\hat i+2\hat j+3\hat k$
$\;\;\;\overrightarrow{OQ}=4\hat i+5\hat j+6\hat k$
$\qquad=4\hat i+5\hat j+6\hat k-(\hat i+2\hat j+3\hat k)$
$\qquad=3\hat i+3\hat j+3\hat k$
Step 2:
Next let us find the unit vector of $\overrightarrow{PQ}$
Magnitude of $\overrightarrow{PQ}=|\overrightarrow{PQ}|=\sqrt{3^2+3^2+3^2}$
$\qquad\qquad\qquad\qquad\;\;\;=\sqrt {27}$
$\qquad\qquad\qquad\qquad\;\;\;=3\sqrt {3}$
Step 3:
Unit vector of $\overrightarrow{PQ}=\large\frac{\overrightarrow{PQ}}{|\overrightarrow{PQ}|}$
$\qquad\qquad\qquad\;\;=\large\frac{3\hat i+3\hat j+3\hat k}{3\sqrt 3}$
$\qquad\qquad\qquad\;\;=\large\frac{3}{3\sqrt 3}$$\hat i+\large\frac{3}{3\sqrt 3}$$\hat j+\large\frac{3}{3\sqrt 3}$$\hat k$
$\qquad\qquad\qquad\;\;=\large\frac{1}{\sqrt 3}$$\hat i+\large\frac{1}{\sqrt 3}$$\hat j+\large\frac{1}{\sqrt 3}$$\hat k$
answered May 17, 2013 by sreemathi.v

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