Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Determinants

Question:

The vector equation of the line joining the points $(-2,-3,-4)$ and $(1,-2,4)$ is :

Options:

$\vec{r}=(-2 \hat{i}-3 \hat{j}-4 \hat{k})+\lambda(\hat{i}-2 \hat{j}+4 \hat{k})$

$\vec{r}=(2 \hat{i}+3 \hat{j}+4 \hat{k})+\lambda(3 \hat{i}-\hat{j}+8 \hat{k})$

$\vec{r}=(-2 \hat{i}-3 \hat{j}-4 \hat{k})+\lambda(3 \hat{i}+\hat{j}+8 \hat{k})$

$\vec{r}=(2 \hat{i}+3 \hat{j}+4 \hat{k})+\lambda(3 \hat{i}+\hat{j}+8 \hat{k})$

Correct Answer:

$\vec{r}=(-2 \hat{i}-3 \hat{j}-4 \hat{k})+\lambda(3 \hat{i}+\hat{j}+8 \hat{k})$

Explanation:

$\vec{x}_1=-2 \hat{i}-3 \hat{j}-4 \hat{k}$

$\vec{x}_2=1 \hat{i}-2 \hat{j}+4 \hat{k}$

so vector parallel to given line is $\left(\vec{x}_2-\vec{x}_1\right)= 3 \hat{i}+\hat{j}+8 \hat{k}$

So equation of lines is

$\vec{r}=\vec{a}+\lambda \vec{b}$ → vector parallel to line

$\vec{r}=(-2 \hat{i}-3 \hat{j}-4 \hat{k})+\lambda(3 \hat{i}+\hat{j}+8 \hat{k})$