Show that lines r=(-i-3 j+4 k)+ (-10 i-j+k) and r=(-10 i-j+k) (-i-3 j+4 k) intersect each other. Find the position vector of their point of intersection.

The position vector of a variable point on the line r=(-i-3 j+4 k)+ (-10 i-j+k) is (-1-10 ) i+(-3- ) j+(4+ ) k The position vector of a variable point on the line r=(-10 i-j+k)+ (-i-3 j+4 k) is (-10-1 ) i+(-1-3 ) j+(1+4 ) k Given lines intersect each other if there exist some values of and for which & (-1-10 ) i+(-3- ) j+(4+ ) k=(-10-1 ) i+(-1-3 ) j+(1+4 ) k & -1-10 =-10-1 ,-3- =-1-3 and 4+ =1+4 & 10 - ,=9, -3 =-2 and -4 =-3 Given lines intersect each other if this system is consistent As | ccc 10 & -1 & 9 1 & -3 & -2 1 & -4 & -3 |=10(9-8)+1(-3+2)+9(-4+3)=10-1-9=0 The system (1) is consistent and lines intersect each other. Solving any two equations in system (1), we get. =1, =1 Substituting this value of in (-1-10 ) i+(-3- ) j+(4+ ) k we get, -11 i-4 j+5 k The position vector of their point of intersection is -11 i-4 j+5 k.

Show that lines r=(-i-3 j+4 k)+ (-10 i-j+k) and r=(-10 i-j+k) (-i…

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Question

Show that lines $\bar{r}=(-\hat{i}-3 \hat{j}+4 \hat{k})+\lambda(-10 \hat{i}-\hat{j}+\hat{k})$ and $\bar{r}=(-10 \hat{i}-\hat{j}+\hat{k}) \mu(-\hat{i}-3 \hat{j}+4 \hat{k})$ intersect each other.
Find the position vector of their point of intersection.

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Answer

The position vector of a variable point on the line $\bar{r}=(-\hat{i}-3 \hat{j}+4 \hat{k})+\lambda(-10 \hat{i}-\hat{j}+\hat{k})$ is $(-1-10 \lambda) \hat{i}+(-3-\lambda) \hat{j}+(4+\lambda) \hat{k}$
The position vector of a variable point on the line $\bar{r}=(-10 \hat{i}-\hat{j}+\hat{k})+\mu(-\hat{i}-3 \hat{j}+4 \hat{k})$ is $(-10-1 \mu) \hat{i}+(-1-3 \mu) \hat{j}+(1+4 \mu) \hat{k}$ Given lines intersect each other if there exist some values of $\lambda$ and $\mu$ for which
$
\begin{aligned}
& (-1-10 \lambda) \hat{i}+(-3-\lambda) \hat{j}+(4+\lambda) \hat{k}=(-10-1 \mu) \hat{i}+(-1-3 \mu) \hat{j}+(1+4 \mu) \hat{k} \\
\therefore \quad & -1-10 \lambda=-10-1 \mu,-3-\lambda=-1-3 \mu \text { and } 4+\lambda=1+4 \mu \\
\therefore \quad & 10 \lambda-\mu,=9, \lambda-3 \mu=-2 \text { and } \lambda-4 \mu=-3
\end{aligned}
$
Given lines intersect each other if this system is consistent
$\operatorname{As}\left|\begin{array}{ccc}10 & -1 & 9 \\ 1 & -3 & -2 \\ 1 & -4 & -3\end{array}\right|=10(9-8)+1(-3+2)+9(-4+3)=10-1-9=0$
$\therefore$ The system (1) is consistent and lines intersect each other.
Solving any two equations in system (1), we get. $\lambda=1, \mu=1$
Substituting this value of $\lambda$ in $(-1-10 \lambda) \hat{i}+(-3-\lambda) \hat{j}+(4+\lambda) \hat{k}$ we get, $-11 \hat{i}-4 \hat{j}+5 \hat{k}$
$\therefore$ The position vector of their point of intersection is $-11 \hat{i}-4 \hat{j}+5 \hat{k}$.