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Show that the lines $\begin{array}{1 1}\overrightarrow{r}=3\hat{i}+2\hat{j}-4\hat{k}+\lambda(\hat{i}+2\hat{j}+2\hat{k});\\\overrightarrow{r}=5\hat{i}-2\hat{j}+\mu(3\hat{i}+2\hat{j}+6\hat{k})\end{array} $ are intersecting. Hence find their point of intersection.

(Note: This question has been split into 2 questions) This question appeared in 65-1,65-2 and 65-3 versions of the paper in 2013.
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1 Answer

If the lines are intersecting then
$(a_2^2-a_1).(b_1 \times b_2)=0$
Therefore $a_2-a_1=2\overrightarrow{i}-4\overrightarrow{j}+4\overrightarrow{k}$
$\overrightarrow{b_1} \times \overrightarrow{b_2}=\begin{bmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ 1 & 2 & 2 \\ 3 & 2 & 6 \end{bmatrix}$
Therefore $(a_2-a_1).(b_1 \times b_2)=(2\overrightarrow{i}-4\overrightarrow{j}+4\overrightarrow{k}).(8\overrightarrow{i}-4\overrightarrow{k})$
Hence they are intersecting lines.
The position vectors of arbitrary points on the given lines are :
$L1 = 3i+2j-4k + \lambda(i+2j+2k)$
$= (3+\lambda)i+(2+2 \lambda)j+(-4+2 \lambda)k$
If they intersect they have a common point.
For some values of $\lambda$ and $\mu$ we must have $(3+\lambda)i+(2+2\lambda)j+(-4+2\lambda)k =(5+3\mu)i+(-2+2\mu)j+6\mu k$
$3+\lambda = 5+3\mu$
$ i.e., \lambda-3\mu=2$
on solving the two equations,
$\mu = -2\; and\; \lambda = -4$
Hence the point of intersection can be got by putting the value of lamda in the first line.
$ 3i+2j-4k+(-4)(i+2j+2k)$
$ -i-6j-12k$
Hence the coordinates of the points of intersection are (-1,-6,-12)
answered Mar 21, 2013 by meena.p
edited Mar 26, 2013 by meena.p

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