The three vertices of a triangle are at A (7,-1,3), B (-3,3,-4) and C(-3,1, 5). Find: (use Dot product) a. The vector projection of RAB on RAC b. The length of the perimeter of the triangle
The three vertices of a triangle are at A (7,-1,3), B (-3,3,-4) and C(-3,1, 5). Find: (use Dot product) a. The vector projection of RAB on RAC b. The length of the perimeter of the triangle
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
Please answer and write correctly and neatly.
The three vertices of a triangle are at A (7,-1,3), B (-3,3,-4) and C(-3,1, 5). Find: (use Dot product)
a. The
b. The length of the perimeter of the triangle
Transcribed Image Text:The three vertices of a triangle are at A(7
To find:
(a) RAB
(b) RAC
(c) The angle BAC
Step3
c)
lution:
)
The vertices of the triangle
A=(7, — 1, 3)= 7i − j + 3k
B=(-3, 3, 4)= −3i + 3j − 4k
C=(−3, 1, 5)= −3i+j+5k
Step4
d)
Where, i, j, and k are the u
Since, the vector RAB is give
RAB=B - A
=(−3i + 3j - 4k)−(7i − j + 3k)
=-3i + 3j - 4k - 7i+j - 3k
=-10i + 4j - 7k
herefore, RAB=-10i +4j - 7k
Since,
A=7i- j + 3k
C=-3i+j+ 5k
The vector RAC is given by
RAC CA
=(-3i+j+5k)-(7i- j + 3k)
=-3i+j+ 5k - 7i+j - 3k
=-10i + 2j + 2k
herefore, RAC=-10i +2j + 2k
Transcribed Image Text:Since,
RAB=-10i + 4j - 7k
RAC -10i + 2j + 2k
RAB RAC (-10i + 4j – 7k).(-10i +2j+2
=(−10)(-10)+(4)(2)+(−7)(2)
=100+ 8 - 14
=108
14
=94
||RAB||= (−10)² +4² + (−7)²
=√/100 + 16 + 49
=√/165
|Rac||=√(−10)² + 2² +2²
100+ 4 + 4
.
-
=√108
=6√3
Let, BAC is the angle betwee
We know that the angle bet
a.b
से || || b ||
Therefore, the angle between
0=cos
OBAC=COS
g, RAB RAC=94, ||RAB||= √165, and ||RAC||
BAC COS
RAB RAC
|| RAB || ||RAC ||
(
94
(√165) (6√3)
calculator we get
Using
OBAC 45. 23793134°
erefore, BAC-45. 23793134°
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