p 112 Q 032 and R 251 are the vertices of triangle Find A

p = (1,-1,2), Q = (0,3,-2), and R = (2,-5,1) are the vertices of triangle. Find A) 2PQ - RP (B)the area of triangle (C) the measure of the angle at vertex P (D) the projection of RP onto QP

Solution

p = (1,-1,2),

Q = (0,3,-2), and

R = (2,-5,1)

Thus, PQ = sqrt [ (1-0)² + (-1-3)² + (2- (-2) )² ] = sqrt (33)

PR = sqrt [ (1-2)² + (-1-(-5))² + (2- 1)² ] = sqrt ( 18 )

Thus, 2PQ - PR = 2sqrt (33) - sqrt(18) = 7.246

c)

For measuring angle at vertex P we need the vectors PQ and PR

PQ = (q - p) = (-1 , 4, -4)

PR = (r - p) = (1 , -4, -1)

cos P = PQ.PR / [ |PQ| |PR| ]

= (-1 * 1) + (-4*4) + (-4*-1) / sqrt (33*18)

cos P = -0.53339

P = 122.235 degrees

b) Area of the triangle = 1/2 * |PQ| . |PR| sin (P)

= 0.5 * sqrt (18*33) * sin (122.235)

= 10.3077

d) Projection of RP onto QP = |RP| cos P

= | RP .QP | / |QP|

RP = ( p - r ) = (-1, 4, 1)

QP = (p - q ) = (1, -4, 4)

thus, projection = ( 1 * -1) + (4*-4) + (1*4) / { sqrt(33) }

= | -2.263 |

= 2.263

Hope this helps. Ask if you have any doubts.