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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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The vector equation of the line through the points $(3,4,-7)$ and $(1,-1,6)$ is ______________.

$\begin{array}{1 1}(A)\;(x-3) \hat i+(y-4) \hat j+(z+7) \hat k=\lambda(-2 \hat i-5 \hat j+14 \hat k) \\(B)\;(x-3) \hat i-(y-4) \hat j+(z+7) \hat k=\lambda(-2 \hat i-5 \hat j+14 \hat k) \\(C)\;(x-3) \hat i+(y-4) \hat j+(z+7) \hat k=\lambda(2 \hat i+5 \hat j+14 \hat k) \\(D)\;(x-3) \hat i-(y-4) \hat j-(z-7) \hat k=\lambda(-2 \hat i-5 \hat j+14 \hat k) \end{array} $

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  • Vector equation of a line passing through two point with points with position vectors $\overrightarrow a$ nad $\overrightarrow b$ is $\overrightarrow r=\overrightarrow a+\lambda (\overrightarrow b-\overrightarrow a)$
Let the given points be $A(3,4,7)$ and $b(1,-1,6)$
Hence $\overrightarrow a=3 \hat i+4 \hat j-7 \hat k$ and $\overrightarrow b= \hat i- \hat j+6 \hat k$
Vector equation of a line passing through two point with points with position vectors $\overrightarrow a$ nad $\overrightarrow b$ is
$\overrightarrow r=\overrightarrow a+\lambda (\overrightarrow b-\overrightarrow a)$-----(1)
Let us first determine $ (\overrightarrow b-\overrightarrow a)$
$ (\overrightarrow b-\overrightarrow a)=( \hat i- \hat j+6 \hat k)-(3 \hat i+4 \hat j-7 \hat k)$
$\qquad\qquad=-2 \hat i-5 \hat j+14 \hat k$
Now substituting for $\overrightarrow a$ and $(\overrightarrow b-\overrightarrow a)$ in equ (1)
$\overrightarrow {r}=3 \hat i+4 \hat j-7 \hat k+\lambda(-2 \hat i-5 \hat j+14 \hat k)$
But $\overrightarrow {r}=x \hat i+y \hat j+z \hat k$
$(x \hat i+y \hat j+z \hat k)=3 \hat i+4 \hat j-7 \hat k+\lambda(-2 \hat i-5 \hat j+14 \hat k)$
=>$(x-3) \hat i+(y-4) \hat j+(z+7) \hat k=\lambda(-2 \hat i-5 \hat j+14 \hat k)$
This is the required equation.
answered Jun 13, 2013 by meena.p
 

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