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  • Find the position vector of a point A in space, so that overline{OA} is inclined at 60⁰ to overline{OX} and 45⁰ to overline{OY} and |overline{OA} |= 10 units.

Find the position vector of a point A in space, so that \overline{OA} is inclined at  60⁰ to \overline{OX} and 45⁰ to \overline{OY} and \left |\overline{OA} \right |= 10 units.

Answers (1)

Given, \overline{OA} is inclined at 600 to and at  \overline{OX} 450 to \overline{OY}

\overline{OA} = 10 units.

We want to find the position vector of point A in space, which is nothing but  \overline{OA}

 We know, there are three axes in space: X, Y, and Z.

Let OA be inclined with OZ at an angle α.

We know, directions cosines are associated by the relation:

l² + m² + n² = 1 ….(i)

In this question, direction cosines are the cosines of the angles inclined by  on  \overline{OA}, \overline{OX}, \overline{OY} and  \overline{OZ}

So,l=\cos 60^{\circ},m=\cos 45^{\circ},n=\cos \alpha

Substituting the values of l, m, and n in equation (i),

\left (\cos 60^{\circ} \right )^{2}+\left (m=\cos 45^{\circ} \right )^{2}+\left (n=\cos \alpha \right )^{2}=1

We know the values of \cos 60^{\circ}and \cos 45^{\circ}, i.e. 1/2 and 1/√2 respectively.

Therefore, we get

\left ( \frac{1}{2} \right )^{2}+\left ( \frac{1}{\sqrt{2}} \right )^{2}+\cos^{2}\alpha =1

\Rightarrow \frac{1}{4}+\frac{1}{2}+\cos^{2}\alpha =1

\Rightarrow \cos^{2}\alpha =1-\frac{1}{4}-\frac{1}{2}

\Rightarrow \cos^{2}\alpha =\frac{4-1-2}{4}

\Rightarrow \cos^{2}\alpha =\frac{1}{4}

\Rightarrow \cos \alpha =\pm \sqrt{\frac{1}{4}}

\Rightarrow \cos \alpha =\pm \frac{1}{2}

So \overrightarrow{OA} is given as

\overrightarrow{OA}=\overrightarrow{OA}\left ( l\hat{i}+m\hat{j}+n\hat{k} \right )..........(ii)

We have,

l = \cos 60^{\circ} = \frac{1}{2}\\ m =\ cos 45^{\circ} = \frac{1}{\sqrt{2}}\\ n = \cos \alpha = \pm \frac{1}{2}\\

Inserting these values of l, m and n in equation (ii),

\overrightarrow{OA}=\left |\overrightarrow{OA} \right |\left ( \frac{1}{2} \hat{i}+\frac{1}{\sqrt{2}}\hat{j}+\frac{1}{2}\hat{k} \right )

Also ,Put | \overrightarrow{OA}| =10

\Rightarrow \overrightarrow{OA}=10\left ( \frac{1}{2} \hat{i}+\frac{1}{\sqrt{2}}\hat{j}+\frac{1}{2}\hat{k} \right )

\Rightarrow \overrightarrow{OA}=10\times \frac{1}{2} \hat{i}+10\times\frac{1}{\sqrt{2}}\hat{j}+10\times\frac{1}{2}\hat{k}

\Rightarrow \overrightarrow{OA}=5i+10\times\frac{\sqrt{2}}{\sqrt{2}}\times+10\times\frac{1}{\sqrt{2}}\hat{j}\pm 5\hat{k}

\Rightarrow \overrightarrow{OA}=5i+\frac{10\times\sqrt{2}}{2}\hat{j}\pm 5\hat{k}

\Rightarrow \overrightarrow{OA}=5i+5\sqrt{2}\hat{j}\pm 5\hat{k}

Thus, the position vector of A in space =5i+5\sqrt{2}\hat{j}\pm 5\hat{k}

Posted by

infoexpert24

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