Explanation
Since line makes an angle of $$\displaystyle \dfrac { \pi }{ 4 } $$ with positive direction of each of $$x$$-axis and $$y$$-axis, therefore $$\displaystyle \alpha =\dfrac { \pi }{ 4 } ,\beta =\dfrac { \pi }{ 4 } $$
We know that,
$$\cos ^{ 2 }{ \alpha } +\cos ^{ 2 }{ \beta } +\cos ^{ 2 }{ \gamma } =1$$
$$\displaystyle \\ \Rightarrow \cos ^{ 2 }{ \dfrac { \pi }{ 4 } } +\cos ^{ 2 }{ \dfrac { \pi }{ 4 } } +\cos ^{ 2 }{ \gamma } =1$$
$$\displaystyle \Rightarrow \dfrac { 1 }{ 2 } +\dfrac { 1 }{ 2 } +\cos ^{ 2 }{ \gamma } =1$$
$$\Rightarrow \cos ^{ 2 }{ \gamma } =0$$
$$\Rightarrow \gamma ={ 90 }^{ 0 }$$
If the direction cosines of a line are $$\left(\displaystyle \dfrac{1}{c},\dfrac{1}{c},\dfrac{1}{c}\right)$$ then $$c=$$______
Let the vector $$v$$ make an angle $$\alpha$$ with each of the three axes, then direction cosine of $$v$$ are
$$<\cos\,\alpha, \cos\,\alpha, \cos\,\alpha>$$
$$\Rightarrow \cos^2\alpha =\dfrac {1}{3}$$
$$\Rightarrow \cos\,\alpha=\pm \dfrac{1}{\sqrt{3}}$$
Hence, direction cosine of $$v$$ are
$$<\dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}}>$$ or
$$<-\dfrac{1}{\sqrt{3}}, -\dfrac{1}{\sqrt{3}}, -\dfrac{1}{\sqrt{3}}>$$
$$6x - 2 = 3y + 1 = 2z - 2$$
$$\dfrac{{x - \dfrac{1}{3}}}{{\dfrac{1}{6}}} = \dfrac{{y + \dfrac {1}{3}}}{{{\dfrac {1}{3}}}} = \dfrac{{z - 1}}{{\dfrac{1}{2}}}$$
$$\overrightarrow r = \overrightarrow a + \lambda \,\overrightarrow b $$
$$a = \left( {{1 \over 3}, - {1 \over 3},1} \right),b = \left( {{1 \over 6},{1 \over 3},{1 \over 2}} \right)$$
$$velue\,e{q^n}\,is\, = \left( {1,2,3} \right)$$
$$\mathop {{\text{ }}r}\limits^ \to = \dfrac{1}{3}\widehat i - \dfrac{1}{3}\widehat j + \widehat k + \lambda \left( {\dfrac{1}{6}\widehat i + \dfrac{1}{3}\widehat j + \dfrac{1}{2}\widehat k} \right)$$
$$\mathop {{\text{ }}r}\limits^ \to = \dfrac{1}{3}\widehat i - \dfrac{1}{3}\widehat j + \widehat k + \lambda \left( {\widehat i + 2\widehat j + 3\widehat k} \right)$$
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