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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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Show that the lines $ \frac{ \large x - 5 }{ \large 7} = \frac{ \large y + 2}{ \large -5} = \frac{ \large z}{ \large 1}$ and $ \frac{ \large x }{ \large 1} = \frac{ \large y}{ \large 2} = \frac{ \large z}{ \large 3}$ are perpendicular to each other.

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Toolbox:
  • Angle between the line is given by
  • $\cos \theta=\bigg|\large\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt {{a_1}^2+{b_1}^2+{c_1}^2}\sqrt {{a_2}^2+{b_2}^2+{c_2}^2}}\bigg|$
Let $L_1$ be the line :$\large\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}$
Hence the Direction cosines of the vector $b_1$ in $L_1$ is $(7,-5,1)$
Let this be $(a_1,b_1,c_1)$
Let $L_2$ be the line :$\large\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$
Hence the Direction cosines of the vector $b_2$ in $L_2$ is $(1,2,3)$
Let this be $(a_2,b_2,c_2)$
we know the angle between the line is
$\cos \theta=\bigg|\large\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt {{a_1}^2+{b_1}^2+{c_1}^2}\sqrt {{a_2}^2+{b_2}^2+{c_2}^2}}\bigg|$
Now substituting for $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$
$\cos \theta=\bigg|\large\frac{7 \times 1+-5 \times 2 +1 \times 3}{\sqrt {7^2+(-5)^2+1^2} \sqrt {1^2+2^2+3^2}}\bigg|$
$\cos \theta=\bigg|\large\frac{7-10+3}{\sqrt {49+25+1} \sqrt {1+4+9}}\bigg|$
$\cos \theta=\bigg|\large\frac{0}{\sqrt {75}\sqrt {14}}\bigg|$
Therefore $\cos \theta=0$
$=> \theta=90^{\circ}$
Hence the lines are perpendicular to each other
answered Jun 5, 2013 by meena.p
 

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