Question

The equation of a line parallel to the vector `3hati + hatj + 2hatk`and passing through the point (4, −3, 7) is ______.

विकल्प

• x = 4t + 3, y = −3t + 1, z = 7t + 2

• x = 3t + 4, y = t + 3, z = 2t + 7

• x = 3t + 4, y = t – 3, z = 2t + 7

• x = 3t + 4, y = −t + 3, z = 2t + 7

Answer 1

The equation of a line parallel to the vector `3hati + hatj + 2hatk`and passing through the point (4, −3, 7) is x = 3t + 4, y = t – 3, z = 2t + 7.

Explanation:

`vecr = veca + λvecb`    ...(equation of line)

According to the question,

`veca = 4hati - 3hatj + 7hatk`    ...(point)

`vecb = (3hati + hatj + 2hatk) + ("as lines are parallel")`

∴ `vecr = veca + tvecb`

= `(4hati - 3hatj + 7hatk) + t(3hati + hatj + 2hatk)`

= `4hati - 3hatj + 7hatk + 3thati + thatj + 2thatk`

= `(4 + 3t)hati + (-3 + t)hatj + (7 + 2t)hatk`

Comparing with `vecr = xhati + yhatj + zhatk`

= `(3t + 4)hati + (–3 + t)hatj + (7 + 2t)hatk`

∴ x = 3t + 4, y = t – 3, z = 2t + 7