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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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Find the position vector of a point A in space such that $\overrightarrow{OA}$ is inclined at $60^{\circ}$ to OX and at $45^{\circ}$ to OY and $\mid \overrightarrow{OA}\mid$=10 units.

$\begin{array}{1 1} \hat i + \sqrt 2 \hat j + \hat k \\ 5\hat i + 5\sqrt 2 \hat j - 5 \hat k \\ 10\hat i + 10\sqrt 2 \hat j + 10 \hat k \\ 5\hat i + 5\sqrt 2 \hat j + 5 \hat k\end{array} $

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  • If $\alpha ,\beta, \gamma $ are the angles made by a vector with the $x,y$ and $z$ axes respectivley, then the sum of the squares of the cosines made by there angles is one $(ie) \cos ^2 \alpha +\cos ^2 \beta +\cos ^2 \gamma=1$
we know that the sum of the squares of the direction cosines is one. $(ie) \cos ^2 \alpha +\cos ^2 \beta +\cos ^2 \gamma=1$
It is given that $\alpha =60 ^{\circ}$ and $\beta=45^\circ$
Therefore $\cos ^2 60 ^{\circ}+\cos ^2 45^{\circ}+\cos ^2 \gamma=1$
$\cos 60=\large\frac{1}{2};$$ \cos 45=\large\frac{1}{\sqrt 2}$
$(ie) \bigg(\large\frac{1}{2}\bigg)^2+\bigg(\frac{1}{\sqrt 2}\bigg)^2+$$\cos ^2 \gamma=1$
$=> \large\frac{1}{4}+\frac{1}{2}$$+\cos ^2 \gamma=1$
On simplifying we get,
$\cos ^2 \gamma=1-\large\frac{1}{4}-\frac{1}{2}$
It is given that the magnitude of the given vector in 10 units
$(ie) |\overrightarrow {OA}|=10$
Therefore $\overrightarrow {OA}=10$
Therefore $\overrightarrow {OA}=10 \bigg(\large\frac{1}{2} \hat i+\frac{1}{\sqrt 2}\hat j+\frac {1}{2}\hat k\bigg)$
$=5 \hat i+5 \sqrt 2 \hat j+5 \hat k$
This is the required position vector
answered Jun 10, 2013 by meena.p

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