Question

Find the equation of the plane passing through the line of intersection of the planes `vecr * (hati + hatj + hatk)` = 1 and `vecr * (2hati + 3hatj - hatk) + 4` = 0 and parallel to x-axis

पर्याय

• `vecr * (hatj - 3hatk) + 6` = 0

• `vecr * (hati - 2hatj - 3hatk)` = 14

• `vecr * (2hati + hatj - hatk)` = – 5

• `vecr * (hati + hatj + hatk) - 1` = 0

Answer 1

`vecr * (hatj - 3hatk) + 6` = 0

Explanation:

The two given plane are

`vecr * (2hati + 3hatj - hatk) + 4` = 0 and `vecr * (hati + hatj + hatk) - 1` = 0

The equation of the plane that passes through the intersection of these planes.

`vecr * (2hati + 3hatj - hatk) + 4 + lambda [vecr * (hati + hatj + hatk) - 1]` = 0

or `vecr * [(2 + lambda)hati + (3 + lambda)hatj + (-1 + lambda)hatk] + 4 - lambda` = 0

⇒ The x-axis is perpendicular to the plane (1)

⇒ Its normal is parallel to the x-axis.

Direction ratios of normal are `2 + lambda, 3 + lambda, -1 + lambda`

Direction ratios of x-axis are 1, 0, 0

The plane's normal and the x-axis are perpendiculars.

`(2 + lambda) xx 1 + (3 + lambda) xx 0 + (-1 + lambda) xx 0` = 0

or `2 + lambda` = 0

∴ `lambda = - 2`

Putting `lambda = - 2` in (1)

`vecr * [(2 - 2)hati + (3 - 2)hatj + (-1 - 2)hatk] + 4 + 2` = 0

∴ The required plane equation is `vecr * (hatj - 3hatk) + 6` = 0.