Match List-I with List-II
| List-I (Straight Lines) | List-II (Direction Ratios) | ||
| (A) | \( \vec{r}=(\hat{\imath}+\hat{\jmath})+\lambda(2\hat{\imath}-\hat{\jmath}+3\hat{k}) \) | (I) | 2,1,−3 |
| (B) | \( \displaystyle\frac{x}{2}=\displaystyle\frac{1-y}{-1}=\displaystyle\frac{z+1}{3} \) | (II) | −2,1,3 |
| (C) | \( \displaystyle\frac{1-2x}{4}=\displaystyle\frac{y+1}{1}=\displaystyle\frac{2z+3}{6} \) | (III) | 2,−1,3 |
| (D) | \( \vec{r}=(2\hat{\imath}-\hat{k})+\lambda(2\hat{\imath}+\hat{\jmath}-3\hat{k}) \) | (IV) | 2,1,3 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(III),(B)-(IV),(C)-(II),(D)-(I)
The direction ratios of a line are the components of the vector multiplied by λ, or the denominators once the line is written as \( \frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c} \), with x, y, z each having coefficient 1.
(A) \( \vec{r}=(\hat{\imath}+\hat{\jmath})+\lambda(2\hat{\imath}-\hat{\jmath}+3\hat{k}) \)
This matches (III) 2, −1, 3, the components of the vector multiplied by λ.
(B) \( \displaystyle\frac{x}{2}=\frac{1-y}{-1}=\frac{z+1}{3} \)
This matches (IV) 2, 1, 3. The middle term is
\( \displaystyle \frac{1-y}{-1}=\frac{y-1}{1} \)
(C) \( \displaystyle\frac{1-2x}{4}=\frac{y+1}{1}=\frac{2z+3}{6} \)
This matches (II) −2, 1, 3. Making the coefficients of x and z equal to 1:
\( \displaystyle \frac{1-2x}{4}=\frac{x-\frac{1}{2}}{-2} \)
\( \displaystyle \frac{2z+3}{6}=\frac{z+\frac{3}{2}}{3} \)
(D) \( \vec{r}=(2\hat{\imath}-\hat{k})+\lambda(2\hat{\imath}+\hat{\jmath}-3\hat{k}) \)
This matches (I) 2, 1, −3, the components of the vector multiplied by λ.
So the matching is (A)-(III), (B)-(IV), (C)-(II), (D)-(I).