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In Exercises 36–41 use the given sets of points to find:

(a) A nonzero vector N perpendicular to the plane determined by the points.

(b) Two unit vectors perpendicular to the plane determined by the points.

(c) The area of the triangle determined by the points.

P(1,6),Q(0,3),R(5,4)

(Hint: Think of the xy-plane as part of 3.)

Short Answer

Expert verified

(a) A nonzero vector N perpendicular to the plane determined by the points are -52k.

(b) Two unit vectors perpendicular to the plane determined by the points are ±k.

(c) The area of the triangle determined by the points is 26.

Step by step solution

01

Step 1. Given Information

In the given exercises use the given sets of points to find:

(a) A nonzero vector N perpendicular to the plane determined by the points.

(b) Two unit vectors perpendicular to the plane determined by the points.

(c) The area of the triangle determined by the points.

The given points areP(1,6),Q(0,3),R(5,4)

02

Part (a) Step 1. firstly finding a nonzero vector N perpendicular to the plane determined by the points.

We haveP(1,6),Q(0,3),R(5,4)

Now,

PQ=(0-1,-3-6,0-0)=(-1,-9,0)PR=(-5-1,4-6,0-0)=(-6,-2,0)

03

Part (a) Step 2. Now finding PQ→×PR→

PQ×PR=ijk-1-90-6-20PQ×PR=i-90-20-j-10-60+k-1-9-6-2PQ×PR=i{(-9)×0-0×(-2)}-j{(-1)×0-0×(-6)}+k{(-1)×(-2)-(-9)×(-6)}PQ×PR=i(0-0)-j(0-0)+k(2-54)PQ×PR=-52k

04

Part (b) Step 1. Now finding two unit vectors perpendicular to the plane determined by the points.

So,

PQ×PR=(0)2+(0)2+(-52)2PQ×PR=±52

Required vector

PQ×PRPQ×PR=-52k±52PQ×PRPQ×PR=±k

05

Part (c) Step 1. Now finding the area of the triangle determined by the points.

AreaABC=12PQ×PRAreaABC=12-52AreaABC=12×52AreaABC=26

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