find the magnitude and direction cosines of vector 3i^- 4 j^+ 5 k^
Answers
Answer:
If
find the magnitude and direction cosines of
(i) a vector + b vector + c vector
(ii) 3a vector - 2b vector + 5c vector
Solution :
(i) a vector + b vector + c vector
= (2i + 3j - 4k) + (3i - 4j - 5k) + (-3i + 2j + 3k)
= (2i + 3i - 3i) + (3j - 4j + 2j) + (-4k - 5k + 3k)
= (2i + j - 6k) vector
|a vector + b vector + c vector| = √22 + 12 + (-6)2
= √(4+1+36) = √41
Direction cosines are (x/r, y/r, z/r)
That is, (2/√41, 1/√41, -6/√41)
Hence magnitude and direction cosines are √41 and (2/√41, 1/√41, -6/√41) respectively.
(ii) 3a vector - 2b vector + 5c vector
Solution :
3a vector = 3(2i+3j-4k)-2(3i-4j-5k)+5(-3i+2j+3k)
= (6-6-15)i + (9+8+10)j+(-12+10+15)k
= -15i + 27j + 13k
|3a vector - 2b vector + 5c vector = √(-15)2 + 272 + 132
= √225 + 729 + 169
= √1123
Direction cosines are (x/r, y/r, z/r)
That is, (15/√1123, 27/√1123, 13/√1123)
Hence magnitude and direction cosines are √1123 and (15/√1123, 27/√1123, 13/√1123) respectively.