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Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
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Find the direction cosines of the vector \( \hat i + 2\hat j + 3\hat k\) .

$\begin{array}{1 1}(A) \bigg( \large\frac{-1}{\sqrt{14}}, \large\frac{2}{\sqrt{14}}, \large\frac{3}{\sqrt{14}}\bigg) \\ (B) \bigg( \large\frac{1}{\sqrt{14}}, \large\frac{2}{\sqrt{14}}, \large\frac{3}{\sqrt{14}}\bigg) \\ (C) \bigg( \large\frac{1}{\sqrt{14}}, \large\frac{-2}{\sqrt{14}}, \large\frac{3}{\sqrt{14}}\bigg) \\(D) \bigg( \large\frac{1}{\sqrt{14}}, \large\frac{2}{\sqrt{14}}, \large\frac{-3}{\sqrt{14}}\bigg) \end{array} $

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1 Answer

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Toolbox:
  • The cosines of the angle made by the vector with the coordinate axes is called direction cosines.
  • Direction cosine (D.C) of vector $ x\hat i + y\hat j + 2\hat k$ is $ \large\frac{x}{\sqrt{x^2+y^2+z^2}}, \large\frac{y}{\sqrt{x^2+y^2+z^2}}, \large\frac{z}{\sqrt{x^2+y^2+z^2}}$
Step 1:
$ Let\:\overrightarrow a = \hat i + 2\hat j + 3\hat k$
The magnitude of $\overrightarrow{a}=|\overrightarrow{a}|=\sqrt{x^2+y^2+z^2}$
$\qquad\qquad\qquad\qquad\qquad=\sqrt{1^2+2^2+3^2}=\sqrt{14}$
step 2:
We know that D.C of vector $ x\hat i + y\hat j + 2\hat k$ is $\bigg( \large\frac{x}{\sqrt{x^2+y^2+z^2}}, \large\frac{y}{\sqrt{x^2+y^2+z^2}}, \large\frac{z}{\sqrt{x^2+y^2+z^2}}\bigg)$
Here $x=1,y=2,z=3$
D.C of $ \overrightarrow a = \bigg( \large\frac{1}{\sqrt{14}}, \large\frac{2}{\sqrt{14}}, \large\frac{3}{\sqrt{14}}\bigg) $
answered May 17, 2013 by sreemathi.v
 

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