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If $P$ is a point in space such that $OP=12$ and $\overrightarrow {OP} $ is inclined with angle $45^{\circ} $ with $OX$ and $60^{\circ}$ with $OY$ then the coordinates of $P$ is ?

$(a)\:(6,6,\pm6\sqrt 2)\:\:\qquad\:(b)\:(6,6\sqrt 2, \pm 6)\:\:\qquad\:(c)\:(6\sqrt 2,6,\pm 6)\:\:\qquad \:(d)\:None\:of\:these.$

1 Answer

  • If a line makes angle $\alpha,\:\beta\:and\:\gamma$ with coordinate axes, then $cos^2\alpha+cos^2\beta+cos^2\gamma=1$
Given that $OP$ makes $45^{\circ}\:and\: 60^{\circ}$ with $x \:axis\:and\:y\:axis$ respectively.
Let it make angle $\gamma$ with $z\:axis$
$\Rightarrow\:cos^2 45^{\circ}+cos^2 60^{\circ}+cos^2 \gamma=1$
$\Rightarrow\: cos^2 \gamma=\large\frac{1}{4}$$\:\:\:or\:\:\:cos\gamma=\pm\large\frac{1}{2}$
If $\overrightarrow {OP}$ makes angle $\alpha,\:\beta\:\gamma$ with coordinate axes, then,
$\hat {OP}=cos\alpha \hat i+cos\beta \hat j+cos\gamma \hat k$
$=\large\frac{1}{\sqrt 2}$$\hat i+\large\frac{1}{2}$$\hat j\pm\large\frac{1}{2}$$\hat k$
But given that $|\overrightarrow {OP}|=12.$
$\Rightarrow\:\overrightarrow {OP}=12(\large\frac{1}{\sqrt 2}$$\hat i+\large\frac{1}{2}$$\hat j\pm\large\frac{1}{2}$$\hat k)=6\sqrt 2\hat i+6\hat j\pm6\hat k$
$\therefore$ the coordinates of $P$ is $ (6\sqrt 2,6,\pm6).$
answered Jan 3, 2014 by rvidyagovindarajan_1

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