Subjective Type

Find the angle between the following pair of lines: (i) $$\cfrac{x-2}{2}=\cfrac{y-1}{5}=\cfrac{z+3}{-3}$$ and $$\cfrac{x+2}{-1}=\cfrac{y-4}{8}=\cfrac{z-5}{4}$$ (ii) $$\cfrac{x}{2}=\cfrac{y}{2}=\cfrac{z}{1}$$ and $$\cfrac{x-5}{4}=\cfrac{y-2}{1}=\cfrac{z-3}{8}$$

Solution

(i)
Let $$\vec { { b }_{ 1 } } $$ and $$\vec { { b }_{ 2 } } $$ be the vectors parallel to the pair of lines,
$$\cfrac{x-2}{2}=\cfrac{y-1}{5}=\cfrac{z+3}{-3}$$ and
$$\cfrac{x+2}{-1}=\cfrac{y-4}{8}=\cfrac{z-5}{4}$$ respectively,
then $$\vec { { b }_{ 1 } } =\left( \hat { 2i } +5\hat { j } -3\hat { k }

\right) $$ and $$\vec { { b }_{ 2 } } =\left( -\hat { i } +8\hat { j }

+4\hat { k } \right) $$
$$\left| \vec { { b }_{ 1 } } \right|= \sqrt

{ { \left( 2 \right) }^{ 2 }+{ \left( 5 \right) }^{ 2 }+{ \left( -3

\right) }^{ 2 } } =\sqrt { 38 } $$
$$\left| \vec { { b }_{ 2 } }

\right| =\sqrt { { \left( -1 \right) }^{ 2 }+{ \left( 8 \right) }^{ 2

}+{ \left( 4 \right) }^{ 2 } } =\sqrt { 81 } =9$$
$$\vec { { b }_{ 1

} } .\vec { { b }_{ 2 } } =\left( \hat { 2i } +5\hat { j } -3\hat { k }

\right) .\left( -\hat { i } +8\hat { j } +4\hat { k } \right) $$
$$=2(-1)+5\times 8+(-3).4=-2+40-12=26$$
The angle $$\theta$$ between the given pair of lines is given by the relation,
$$\Rightarrow \cos

{ \theta } =\left| \cfrac { \vec { { b }_{ 1 } } .\vec { { b }_{ 2 } } }{

\left| \vec { { b }_{ 1 } } \right| \left| \vec { { b }_{ 2 } }

\right| } \right| $$
$$\Rightarrow$$ $$\cos { \theta } =\cfrac { 26 }{ 9\sqrt { 38 } } $$
$$\Rightarrow$$ $$\theta=\cos ^{ -1 }{ \left( \cfrac { 26 }{ 9\sqrt { 38 } } \right) } $$

(ii)
We have,
$$\vec { { b }_{ 1 } } =\left( \hat { 2i } +2\hat { j } +\hat { k } \right) $$
$$\vec { { b }_{ 2 } } =\left( 4\hat { i } +\hat { j } +8\hat { k } \right) $$
$$\therefore$$ $$\left| \vec { { b }_{ 1 } } \right| =\sqrt { { \left( 2 \right) }^{

2 }+{ \left( 2 \right) }^{ 2 }+{ \left( 1 \right) }^{ 2 } } =\sqrt { 9

} =3$$
$$\left| \vec { { b }_{ 2 } } \right| =\sqrt { { \left( 4

\right) }^{ 2 }+{ \left( 1 \right) }^{ 2 }+{ \left( 8 \right) }^{ 2 }

} =\sqrt { 81 } =9$$
$$\vec { { b }_{ 1 } } .\vec { { b }_{ 2 } }

=\left( \hat { 2i } +2\hat { j } +\hat { k } \right) .\left( 4\hat { i }

+\hat { j } +8\hat { k } \right) $$
$$=2\times 4+2\times 1+1\times 8=8+2+8=18$$
If $$\theta$$ is the angle between the given pair of lines, then $$\cos { \theta }

=\left| \cfrac { \vec { { b }_{ 1 } } .\vec { { b }_{ 2 } } }{ \left|

\vec { { b }_{ 1 } } \right| \left| \vec { { b }_{ 2 } } \right| }

\right| $$
$$\Rightarrow$$ $$\cos { \theta } =\cfrac { 18 }{ 3\times 9 } =\cfrac { 2 }{ 3 } $$
$$\Rightarrow$$ $$\theta =\cos ^{ -1 }{ \left( \cfrac { 2 }{ 3 } \right) } $$

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