Vectors

[brinlin]

Find a vector that is perpendicular to the plane passing through the points P (1, 2, 3), Q (2, 3, 1), and R (3, 1, 2).

 

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[HallsofIvy]

Well, first, find the equation of the plane!

Any plane can be written \(\displaystyle A(x- x_0)+ B(y- y_0)+ C(z- z_0)= 0\) for some A, B, and C and \(\displaystyle (x_0, y_0, z_0)\) a point in the plane.

Since the plane contains the point (1, 2, 3) we can take \(\displaystyle (x_0, y_0, z_0)= (1, 2, 3)\) and write the plane A(x- 1)+ B(y- 2)+ C(z- 3)= 0.

We can further write that as \(\displaystyle (Ai+ Bj+ Ck)\cdot ((x-1)i+ (y- 2)j+ (z- 3)k)= 0\). Since (x-1)i+ (y- 2)j+ (z- 3)k is a vector in the plane, Ai+Bj+ Ck is a vector perpendicular to the plane.

Since the plane contains the point (2, 3, 1) we must have
A(2- 1)+ B(3- 2)+ C(1- 3)= A+ B- 2C= 0.

Since the plane contains the point (3, 2, 1) we must have
A(3- 1)+ B(2- 2)+ C(1- 3)= 2A- 2C= 0.

From 2A- 2C= 0, C= A so A+ B- 2C= A+ B- 2A= B- A= 0 so B= A.

We can write the equation of the plane as A(x- 1)+ A(y- 2)+ A(z- 3)= 0 and the vector Ai+ Aj+ Ak is perpendicular to the plane for any number A. In particular, taking A= 1, i+ j+ k is perpendicular to the plane.

 

[Dr.Peterson]

Find a vector that is perpendicular to the plane passing through the points P (1, 2, 3), Q (2, 3, 1), and R (3, 1, 2).

Have you learned about cross products (vector products)? If so, think about vectors PQ and PR, for example.

 

[HallsofIvy]

Dr. Peterson's suggestion was, as usual, better than mine. The vectors PQ and PR lie in the plane and their cross product is perpendicular to the plane.

 

[pka]

Find a vector that is perpendicular to the plane passing through the points P (1, 2, 3), Q (2, 3, 1), and R (3, 1, 2).

We have [imath]\overrightarrow {PQ} = \left\langle {1,1, - 2} \right\rangle \;\& \;\overrightarrow {PR} = \left\langle {2, - 1, - 1} \right\rangle[/imath]
What is [imath]\overrightarrow {PQ} = \left\langle {1,1, - 2} \right\rangle \times\overrightarrow {PR} = \left\langle {2, - 1, - 1} \right\rangle=?[/imath]

 

[HallsofIvy]

Since this has been up here a week (and I just can't resist)

\(\displaystyle \vec{PQ}\times \vec{PR}= \begin{vmatrix}\vec{i} & \vec{j} & \vec{k} \\ 1 & 1 & -2 \\ 2 & -1 & -1 \end{vmatrix}= \vec{i}\begin{vmatrix}1 & -2 \\ -1 & -1 \end{vmatrix}- \vec{j}\begin{vmatrix}1 & -2 \\ 2 & -1\end{vmatrix}+ \vec{k}\begin{vmatrix}1 & 1 \\ 2 & -1\end{vmatrix}\)
\(\displaystyle = \vec{i}(-1- 2)- \vec{j}(-1+ 2)+ \vec{k}(-1- 2)= -3\vec{i}- \vec{j}- 3\vec{k}\).

Of course, swapping \(\displaystyle \vec{PQ}\) and \(\displaystyle \vec{PR}\) in the determinant would change the sign, giving \(\displaystyle 3\vec{i}+ \vec{j}+ 3\vec{k}\), in the opposite direction, but still perpendiculr to the plane.