Find the distance of the point (−1, −5, −10) from the point of intersection of the line and the plane.

Open in App

Solution

The equation of the given line is

The equation of the given plane is

Substituting the value of from equation (1) in equation (2), we obtain

Substituting this value in equation (1), we obtain the equation of the line as

This means that the position vector of the point of intersection of the line and the plane is

This shows that the point of intersection of the given line and plane is given by the coordinates, (2, −1, 2). The point is (−1, −5, −10).

The distance d between the points, (2, −1, 2) and (−1, −5, −10), is

Suggest Corrections

Similar questions

(- 1, - 5, - 10)

\to r = 2^i -^j + 2^k + λ (3^i + 4^j + 2^k)

\to r \cdot (^i -^j +^k) = 5

(2, 12, 5)

\to r = 2^i - 4^j + 2^k + λ (3^i + 4^j + 2^k)

\to r \cdot (^i - 2^j +^k) = 0.

\to r \cdot (3^i -^j +^k) = 1

\to r \cdot (^i + 4^j - 2^k) = 2

¯ r = 2^i - 4^j + 2^k + λ (3^i + 4^j + 2^k)

¯ r \cdot (^i - 2^j +^k) = 0

\to r = 2^i - 4^j + 2^k + λ (3^i + 4^j + 2^k)

\to r . (^i - 2^j +^k) = 0

Join BYJU'S Learning Program