Match List-I with List-II.
| List-I | List-II | ||
| (A) | The equation of the line passing through the points (-1, 0, 2) and (3, 4, 6) is | (I) | $\vec{r}=(-\hat{i}+2\hat{k})+\lambda (4\hat{i}+4\hat{j}+4\hat{k})$ |
| (B) | The vector equation of the line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{4-z}{5}$ is | (II) | $\vec{r}=(\hat{i}+3\hat{j}+2\hat{k})+\lambda (2\hat{i}+\hat{j}+3\hat{k})$ |
| (C) | The equation of line passing through (1, 3, 2) and parallel to $\frac{x+1}{2}=\frac{y-1}{1}=\frac{z+1}{3}$ is | (III) | $\vec{r}=(-\hat{i}+2\hat{j}+3\hat{k})+\lambda (2\hat{i}+\hat{j}+4\hat{k})$ |
| (D) | The equation of the line along $2\hat{i}+\hat{j}+4\hat{k}$ and passing through (-1, 2, 3) is | (IV) | $\vec{r}=(\hat{i}-\hat{j}+4\hat{k})+\lambda (2\hat{i}+3\hat{j}-5\hat{k})$ |
Where $\lambda $ is an arbitrary constant.
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(I),(B)-(IV),(C)-(II),(D)-(III)
(A) equation → $\vec r=\vec a+λ(\vec b-\vec a)$
$\vec a=-\hat i+2\hat k,\vec b=3\hat i+4\hat j+6\hat k$
$⇒\vec{r}=-\hat{i}+2\hat{k})+\lambda (4\hat{i}+4\hat{j}+4\hat{k})$ (I)
(B) Point $(\hat i-\hat j+4\hat k)$
$\vec v$ || line $(2\hat i+3\hat j-5\hat k)$
$⇒\vec{r}=(\hat{i}-\hat{j}+4\hat{k})+\lambda (2\hat{i}+3\hat{j}-5\hat{k})$ (IV)
(C) point $(\hat i-\hat j+4\hat k)$
line || $\vec v$ || direction ratios of given line
$\vec v=2\hat i+\hat j+3\hat k$
$⇒\vec{r}=(\hat{i}+3\hat{j}+2\hat{k})+\lambda (2\hat{i}+\hat{j}+3\hat{k})$ (II)
(D) point $(-\hat i+2\hat j+3\hat k)$
$\vec v$ || line: $\vec v=2\hat i+\hat j+4\hat k$
$\vec{r}=(-\hat{i}+2\hat{j}+3\hat{k})+\lambda (2\hat{i}+\hat{j}+4\hat{k})$ (III)