Find a vector perpendicular to the plane that passes through the points:p(1, 4, 6), Q(-2, 5, -1) and R(1. -1, 1)

Answers

This answer is not selected as best answer. This answer may not be sufficient for exam.

Your limit has been exceed. We have implemented this system because, We got difficulty on managing our servers. Please donate some amount to remove this limit.

Quota: 0 / 30

Donate

Solution

First, We will find vector form of (1, 4, 6) (-2, 5, -1) and (1, 4, 6) (1, -1, 1). As all three points are in the plane, so will each of those vectors

\(\vec{u_{1}}\) = (1, 4, 6) – (-2, 5, -1) = (3, -1, 7)

\(\vec{u_{2}}\) = (1, 4, 6) – (1, -1, 1) = (0, 5, 5)

If a vector is perpendicular to two vectors in a plane, it must be perpendicular to the plane itself. As the cross product of two vectors produces a vector perpendicular to both, we will use the cross product of \(\vec{u_{1}}\) and \(\vec{u_{2}}\) to find a vector \(\vec{u}\) perpendicular to the plane containing them.

\(\Rightarrow \vec{u} = \vec{u_{2}} \times \vec{u_{2}}\)

\(\Rightarrow \vec{u} = \begin{vmatrix}
\vec{i} & \vec{j} & \vec{k}\\
3 & -1 & 7\\
0 & 5 & 5
\end{vmatrix}\)

\(\Rightarrow 40\vec{i} – 15\vec{j} + 15\vec{k}\)

 

If you found any type of error on the answer then please mention on the comment or submit your new answer.
Leave your Answer:

Click here to submit your answer.

Discussion
0 Comments
  Loading . . .