The point of intersection of vector lines $(7\hat{i}+10\hat{j}+13\hat{k})+k_1(2\hat{i}+3\hat{j}+4\hat{k})$ and $(3\hat{i}+5\hat{j}+7\hat{k})+k_2(\hat{i}+2\hat{j}+3\hat{k})$ is given by

Given:

The two vector lines $(7\hat{i}+10\hat{j}+13\hat{k})+k_1(2\hat{i}+3\hat{j}+4\hat{k})$ and $(3\hat{i}+5\hat{j}+7\hat{k})+k_2(\hat{i}+2\hat{j}+3\hat{k})$

Concept: 

The point of intersection of the two vector lines.

Solution:

Arrange the given vector lines  $(7\hat{i}+10\hat{j}+13\hat{k})+k_1(2\hat{i}+3\hat{j}+4\hat{k})$ let say L₁ and $(3\hat{i}+5\hat{j}+7\hat{k})+k_2(\hat{i}+2\hat{j}+3\hat{k})$  let say L₂ as follows,

$L_1:\overrightarrow{r}=\left(\begin{array}{c}7\\ 10\\ 13\end{array}\right)+\left(\begin{array}{c}2\\ 3\\ 4\end{array}\right)k_1$          (A)

$L_2:\overrightarrow{r}=\left(\begin{array}{c}3\\ 5\\ 7\end{array}\right)+\left(\begin{array}{c}1\\ 2\\ 3\end{array}\right)k_2$          (B)

Write down the parametric equation for L1  and L2 as follows,

For L1 ,

$$\begin{gathered} x=7+2k_1 \\ y=10+3k_1 \\ z=13+4k_1 \end{gathered}$$

For  L2 ,

$$\begin{gathered} x=3+k_2 \\ y=5+2k_2 \\ z=7+3k_2 \end{gathered}$$

Equating values of x, y, and z from the parametric equation for L1  and L2 we will get

$$\begin{gathered} 7+2k_1=3+k_2 \\ \Rightarrow 2k_1-k_2=-4 \end{gathered}$$          (1)

$$\begin{gathered} 10+3k_1=5+2k_2 \\ \Rightarrow 3k_1-2k_2=-5 \end{gathered}$$          (2)

$$\begin{gathered} 13+4k_1=7+3k_2 \\ \Rightarrow 4k_1-3k_2=-10 \end{gathered}$$          (3)

Solving equations (1), (2), and (3) for k₁ and k₂ we will get,

k1 = -3, k2 = -2

Substituting the value of k1 for (A) and k2 for (B) either we will get

For (A)

  $$\begin{gathered} (7\hat{i}+10\hat{j}+13\hat{k})-3(2\hat{i}+3\hat{j}+4\hat{k}) \\ =\hat{i}+\hat{j}+\hat{k} \end{gathered}$$