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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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Prove that the line through $A(0,-1,-1)$ and $B(4,5,1)$ intersects the line through $C(3,9,4)$ and $D(-4,4,4)$

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  • Equation of a line passing through the points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is $\large \frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$
Let the given points be $A(0,-1,-1)$ and $ (4,5,1)$
We know that the equation of the line passing through two points is given by $\large \frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$
Therefore Equation of line through the point AB is
is $ \large\frac{x-0}{4-0}=\large\frac{y+1}{5+1}=\large\frac{z+1}{1+1}$
$ \large\frac{x}{4}=\large\frac{y+1}{6}=\large\frac{z+1}{2}$
or $ \large\frac{x}{2}=\large\frac{y+1}{3}=\large\frac{z+1}{1}$
Let this line $L_1$ be equal to $\lambda$
$(ie) \large\frac{x}{2}=\large\frac{y+1}{3}=\large\frac{z+1}{1}$$=\lambda$
Equation of line through $C(3,9,4)\: and\: D(-4,4,4)$ is
$ \large\frac{x-3}{-4-3}=\large\frac{y-9}{4-9}=\large\frac{z-4}{4-4}$
$ \large\frac{x-3}{-7}=\large\frac{y-9}{-5}=\large\frac{z-4}{0}$
or $ \large\frac{x-3}{7}=\large\frac{y-9}{5}=\large\frac{z-4}{0}$
Let this line $L_2$ be equal to $\mu$
$ \large\frac{x-3}{7}=\large\frac{y-9}{5}=\large\frac{z-4}{0}=\mu$
Hence any point on line $L_1$ is given by
$P(2\lambda,3\lambda-1,\lambda-1)$
Any point on line $L_2$ is given by
$Q(7\mu+3,5\mu+9,0\mu+4)$
If the lines intersect then the coordinates of $P=Q$
Hence $ 2\lambda=7\mu+3$
$ \qquad\qquad\Rightarrow 2\lambda-7\mu=3$-----(1)
$\qquad\qquad \Rightarrow \lambda-1=4$
$\qquad\qquad\Rightarrow \lambda = 5$-----(2)
$3 \lambda-1=5\mu+9=>3 \lambda-5 \mu=10$
Substituting for $\lambda$ in equ (1) we get
$10-2\mu=3$
$-2 \times (5)-7 \mu =3$
$-7 \mu=-7$
$=>\mu=1$
Substituting this in equ (3)
$3(5)-5(1)=10$
Hence the values of $\lambda$ and $\mu$ satisfies the above equation.
Therefore the lines intersect each other
answered Jun 12, 2013 by meena.p
 

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