Given a line L: \( \displaystyle\frac{x}{1}=\displaystyle\frac{y-1}{2}=\displaystyle\frac{z-2}{3} \), then
Match the LIST-I with LIST-II
| LIST-I | LIST-II | ||
| A. | The foot of perpendicular from point (1, 6, 3) on line L | I. | \( 2\sqrt{13} \) |
| B. | Perpendicular distance from point (1, 6, 3) to line L | II. | (1, 0, 7) |
| C. | Image of point (1, 6, 3) with respect to line L | III. | (1, 3, 5) |
| D. | Distance between points (1, 6, 3) and (1, 0, 7) | IV. | \( \sqrt{13} \) |
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
Let P = (1, 6, 3). Any point of L is \( N(\lambda,\,1+2\lambda,\,2+3\lambda) \).
A. The foot of perpendicular from point (1, 6, 3) on line L
This matches III. (1, 3, 5).
For the foot N, PN is perpendicular to the direction (1, 2, 3) of L:
\( \displaystyle \overrightarrow{PN}=(\lambda-1,\ 2\lambda-5,\ 3\lambda-1) \)
\( \displaystyle (\lambda-1)+2(2\lambda-5)+3(3\lambda-1)=0 \)
\( \displaystyle \Rightarrow\ 14\lambda-14=0 \)
\( \displaystyle \Rightarrow\ \lambda=1 \)
\( \displaystyle N=(1,3,5) \)
B. Perpendicular distance from point (1, 6, 3) to line L
This matches IV. \( \sqrt{13} \).
\( \displaystyle PN=\sqrt{0^2+(-3)^2+2^2}=\sqrt{13} \)
C. Image of point (1, 6, 3) with respect to line L
This matches II. (1, 0, 7).
The foot N is the mid-point of P and its image P':
\( \displaystyle P'=2N-P \)
\( \displaystyle =(2-1,\ 6-6,\ 10-3) \)
\( \displaystyle =(1,0,7) \)
D. Distance between points (1, 6, 3) and (1, 0, 7)
This matches I. \( 2\sqrt{13} \).
\( \displaystyle \sqrt{0^2+6^2+4^2}=\sqrt{52} \)
\( \displaystyle =2\sqrt{13} \)
So the matching is A-III, B-IV, C-II, D-I.