Question

Consider the points below.
P(2, 0, 2),  Q(−2, 1, 3),  R(4, 2, 4)
(a) Find a nonzero vector orthogonal to the plane through the points
P, Q,
and R.

Answer 1

Answer:

a) The nonzero vector orthogonal to the plane through the points P, Q and R is (0, 10, -10).

b) The area of the triangle PQR is $5\sqrt{2}$.

Step-by-step explanation:

The statement is incomplete. The complete question will be presented below:

Consider the points below: P(2, 0, 2),  Q(−2, 1, 3),  R(4, 2, 4)

(a) Find a nonzero vector orthogonal to the plane through the points

P, Q,  and R. (b) Find the area of the triangle PQR.

a) Let $P(x,y,z) = (2,0,2)$, $Q(x,y,z) = (-2,1,3)$ and $R(x,y,z) = (4,2,4)$. First, we construct vectors PQ and PR by using the following formulas:

$\overrightarrow{PQ} = Q(x,y,z) -P(x,y,z)$ (1)

$\overrightarrow{PQ} = (-2,1,3)-(2,0,2)$

$\overrightarrow{PQ} = (-4,1,1)$

$\overrightarrow{PR} = R(x,y,z) -P(x,y,z)$ (2)

$\overrightarrow{PR} = (4,2,4)-(2,0,2)$

$\overrightarrow{PR} = (2,2,2)$

Then, we can find the nonzero vector orthogonal to the plane by means of the following cross product:

$\vec S = \overrightarrow{PQ}\,\times\,\overrightarrow{PR}$

Which can be solved by the following determinant (Law of Sarrus):

$\vec S = \left|\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\-4&1&1\\2&2&2\end{array} \right |$

$\vec S = (2-2,2+8, -8-2)$

$\vec S = (0, 10,-10)$

The nonzero vector orthogonal to the plane through the points P, Q and R is (0, 10, -10).

b) The area of the triangle is determined by the following formula:

$A = \frac{1}{2}\cdot \|\overrightarrow{PQ}\,\times\,\overrightarrow{PR} \|$

$A = \frac{1}{2}\cdot \|\vec S\|$ (3)

$A = \frac{1}{2}\cdot \sqrt{0^{2}+10^{2}+(-10)^{2}}$

$A = 5\sqrt{2}$

The area of the triangle PQR is $5\sqrt{2}$.