CBSE Questions for Class 12 Commerce Maths Three Dimensional Geometry Quiz 2 - MCQExams.com

 $$D.\mathrm{C}$$ of the line are $$\displaystyle \dfrac{1}{\sqrt{3}}$$ , $$\displaystyle \dfrac{1}{\sqrt{3}}$$ , $$\displaystyle \dfrac{1}{\sqrt{3}}$$

Any point on the line at a distance $$t$$ from $${P}(2, -1,2)$$ is $$\left (2+\displaystyle \dfrac{{t}}{\sqrt{3}}, -1+\dfrac{{t}}{\sqrt{3}}, 2+\dfrac{{t}}{\sqrt{3}}\right)$$ 

which lies on $$2x+y+z=9$$.

$$\Rightarrow t=\sqrt 3$$

Let $$\overrightarrow { { m }_{ 1 } } $$ and $$\overrightarrow { { m }_{ 2 } }$$ be vectors parallel to the two given lines.

Then, angle between the two given lines is same as the angle between $$\overrightarrow { { m }_{ 1 } } $$ and $$\overrightarrow { { m }_{ 2 } }$$.

$$\overrightarrow { { m }_{ 1 } } =$$ Vector parallel to the line with direction ratios $$(1,1,2)=i+j+2k$$ and

$$\overrightarrow { { m }_{ 2 } } =$$ Vector parallel to the line with direction ratio $$\left( \left( \sqrt { 3 } -1 \right) ,\left( -\sqrt { 3 } -1 \right) ,4 \right) =\left( \sqrt { 3 } -1 \right) i+\left( -\sqrt { 3 } -1 \right) j+4k$$

Let $$\theta$$ be the angle between the given lines. 

Then $$\displaystyle \cos { \theta  } =\dfrac { \overrightarrow { { m }_{ 1 } } .\overrightarrow { { m }_{ 2 } }  }{ \left| \overrightarrow { { m }_{ 1 } }  \right| \left| \overrightarrow { { m }_{ 2 } }  \right|  } =\dfrac { \left( \sqrt { 3 } -1 \right) -\left( \sqrt { 3 } +1 \right) +8 }{ \sqrt { 1+1+4 } \sqrt { { \left( \sqrt { 3 } -1 \right)  }^{ 2 }+{ \left( \sqrt { 3 } +1 \right)  }^{ 2 }+16 }  } $$

$$\Rightarrow \displaystyle \cos { \theta  } =\dfrac { 6 }{ \sqrt { 6 } \sqrt { 24 }  } =\dfrac { 1 }{ 2 } $$

$$\Rightarrow \theta =\dfrac { \pi  }{ 3 } $$

Let $$\theta$$ be the angle between the given lines. 

The given lines are parallel to the vector. 

$$\overrightarrow { { b }_{ 1 } } =1+2j+2k$$ and $$\overrightarrow { { b }_{ 2 } } =3i+2j+6k$$ respectively.

So, the angle $$\theta$$ between them is given by

$$\displaystyle \cos { \theta  } =\dfrac { \overrightarrow { { b }_{ 1 } } .\overrightarrow { { b }_{ 2 } }  }{ \left| \overrightarrow { { b }_{ 1 } }  \right| \left| \overrightarrow { { b }_{ 1 } }  \right|  } =\dfrac { \left( i+2j+2k \right) \left( 3i+2j+6k \right)  }{ \left| i+2j+2k \right| \left| 3i+2j+6k \right|  } $$

$$\displaystyle =\dfrac { 3+4+12 }{ \sqrt { 1+4+4 } \sqrt { 9+4+36 }  } =\dfrac { 19 }{ 21 } $$

$$\Rightarrow \theta =\cos ^{ -1 }{ \left( \dfrac { 19 }{ 21 }  \right)  } $$

Since, the plane is perpendicular to the z axis, $$a = b=0$$

The plane passes through (2,-3,4).

So $$d = 4c$$

$$ \dfrac{a+b+c}{d} = \dfrac{c}{4c} = \dfrac{1}{4} $$

Let $$\vec{a_1}$$ and $$\vec{a_2} $$ be unit vectors along the lines. 

$$\vec{a_1} = l_1 \hat{i} + m_1 \hat{j} + n_1 \hat{k}$$

$$\vec{a_2} = l_2 \hat{i} + m_2 \hat{j} + n_2 \hat{k}$$

$$\Rightarrow l_1^2+m_1^2+n_1^2 = 1$$ and $$l_2^2+m_2^2+n_2^2 = 1$$

Consider the dot product of $$\vec{a_1} $$ and $$\vec{a_2} $$

$$\vec{a_1}.\vec{a_2} = (l_1l_2)+(m_1m_2)+(n_1n_2) $$

$$\Rightarrow \cos \theta = (l_1l_2)+(m_1m_2)+(n_1n_2) $$

$$S = \displaystyle \dfrac{\Sigma(l_{1}+l_{2})^{2}}{4\cos^{2}(\dfrac{\theta}{2})}+\dfrac{\Sigma(l_{{I}}-l_{2})^{2}}{4\sin^{2}(\dfrac{\theta}{2})}$$

$$\Rightarrow S = \dfrac{ 2 + 2 (l_1l_2 + m_1m_2+ n_1n_2) } {4 \cos^2{\dfrac{\theta}{2}}} + \dfrac{ 2 - 2 (l_1l_2 + m_1m_2+ n_1n_2) } {4 \sin^2{\dfrac{\theta}{2}}} $$

$$\Rightarrow S = \dfrac{2+2\cos\theta}{ 4 \cos^2{\dfrac{\theta}{2}}} + \dfrac{2-2\cos\theta}{4 \sin^2{\dfrac{\theta}{2}}}$$

Now, $$1+\cos\theta = 2\cos^2\dfrac{\theta}2$$ and $$1-\cos\theta = 2\sin^2\dfrac{\theta}2$$

$$\Rightarrow S = \dfrac22 + \dfrac22$$

$$\Rightarrow S =2 $$