Question

Consider the points below. P(0, −4, 0), Q(5, 1, −3), R(5, 2, 1) (a) Find a nonzero vector orthogonal to the plane through the points P, Q, and R. Correct: Your answer is correct. (b) Find the area of the triangle PQR.

Answer 1

Answer:

a. (23,-20,-5)

b. A=15.4

Step-by-step explanation:

first we find two vectors on the plan with length p q and PR.

We have

P Q=<5,1,-3> - <0,-4,0>

P Q=<5-0,1-(-4),-3-0>

P Q=(5,5,-3)

PR=<5,2,1> - <0,-4,0>

PR=<5-0,2-(-4),1-0>

PR=(5,6,1)

The cross product of the two determined vector will produce an orthogonal to the plane

the cross product of two vectors is expressed as

(a,b,c) x (d,e,f)=(bf-c e, c d-a f, a e-b d)

going by the expansion above we have

P Q * PR =(5,5,-3) X (5,6,1) =(5-(-18), -15-5, 30-25)

P Q * PR =(23,-20,-5).

b. To determine the area, we use also use the cross product of the two formed vectors i.e

$A=\frac{1}{2}|PQxPR|$

$A=\frac{1}{2}|(5,5,-3)x(5,6,1)|\\A=\frac{1}{2}|(23,-20,-5)|\\\\A=\frac{1}{2}\sqrt{23^{2}+20^{2}+5^{2}} \\A=\frac{1}{2}(30.9)\\A=15.4 unit square$