MCQ
The vector equation r = i − 2j − k + t(6j − k) represents a straight line passing through the points:
- A(0, 6, −1) and (1, −2, −1)
- B(0, 6, −1) and (−1, −4, −2)
- C(1, −2, −1) and (1, 4, −2)
- D(1, −2, −1) and (0, −6, 1)
Solution:
Cartesian representation of the given line is,
$\frac{\text{x}-1}{0}=\frac{\text{y}+2}{6}=\frac{\text{z}+1}{-1}=\text{t}$
So any point on the given line is of the form (1, 6t − 2, − t − 1) where t can be any real numbers
So for t = 0 and 1 the corresponding points are (1, −2, −1) and (1, 4, −2)
You can check other options does not satisfy above point for any t.
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| X = xi | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| P(X = Xi) | 0 | 2p | 2p | 3p | p2 | 2p2 | 7p2 | 2p |