(a) Find the area of the triangle having vertices A(1, 0, 1), B(0, 2, 3), and C (2, 1, 0). (b) Use the result of part (a) to find the length of the altitude from vertex C to side AB.

Question

(a) Find the area of the triangle having vertices A(1, 0, 1), B(0, 2, 3), and C (2, 1, 0).   (b) Use the result of part (a) to find the length of the altitude from vertex C to side AB.

Elois Puryear · Accepted Answer

Step 1  Given:  A(1,0,1) , B(0,2,3)  ,C(2,1,0)  Step 2  AB→=B→−A→=(0,2,3)−(1,0,1)=(1,−2,−2)  BC→=C→−B→=(2,1,0)−(0,2,3)=(−2,1,3)  AB→×BC→=[ijk1−2−2−213]  =i(−6+2)−j(3−4)+k(1−4)  =−4i+j−3k  Area of triangle =12|AB→×BC→|  =12(−4)2+(1)2+(−3)2  =1216+1+9  =262  Step 3  Distance between AB=(x2−x1)2+(y2−y1)2+(z2−z1)2  =(0−1)2+(2−0)2+(3−1)2  =1+4+4  =3  Area of triangle =262  12×base×height=262  ⇒12×3×height=262  ⇒height=263

Neil Dismukes · Accepted Answer

(a) We have AB→=B−A=(0,2,3)−(1,0,1)=(−1,2,2)  and AC→=C−A=(2,1,0)−(1,0,1)=(1,1,−1)  Then we have AB→×AC→=(−1,2,2)×(1,1,−1)  =([221−1],−[−121−1],[−1211])  =(−4,1,−3)  Hence the area of the triangle with the vertices A, B and C is given by:  12||AB→×AC→||=12||−41−3||=12(−4)2+12+(−3)2=262  (one half of the area of the parallelogram determined by AB→ and AC→).  (b) Let h be the length of the altitude from C to AB. Then the area of the triangle with the vertices A, B and C is 12||AB→||h. Therefore, by using part (a) we get 12||AB→||h=262  Also, we have ||AB→||=||−122||=(−1)2+22+22=9=3  This implies h=26||AB→||=263

user_27qwe · Accepted Answer

(a) Following example 10,A=12||AB→×AC→||AB→=(0,2,3)−(1,0,1)=(−1,2,2)AC→=(2,1,0)−(1,0,1)=(1,1,−1)AB→×AC→=[ijk−12211−1]=(−2−2)i−(1−2)j+(−1−2)k=−4i+j−3kA=12(16+1+9)=12(26)=262(b) area of△ABC=12||AB→||hh=area of△ABC12||AB→||=262121+4+4=262129=26232=263

(a) Find the area of the triangle having vertices A(1,

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