Question

1
À = 3i-j-4k
B =-2i+4j-3k
C = I+2j-k

find a unit vector parallel to 3A-2B+4C​

Answer 1

Answer:

If a be any vector and we are to find the unit vector parallel to a, we actually find the unit vector along a, which is determined by

a / a , where a = | a |

Solution:

The given vectors are

a = 3i - j - 4k = (3, - 1, - 4) ,

b = - 2i + 4j - 3k = (- 2, 4, - 3) and

c = i + 2j - k = (1, 2, -1)

Then p = (3a - 2b + 4c)

= 3 (3, - 1, - 4) - 2 (- 2, 4, - 3) + 4 (1, 2, - 1)

= (9, - 3, - 12) + (4, - 8, 6) + (4, 8, - 4)

= (9 + 4 + 4, - 3 - 8 + 8, - 12 + 6 - 4)

= (17, - 3, - 10)

= 17i - 3j - 10k

Now the unit vector parallel to p is the unit vector along its direction; determined by

p / p , where p = | p |

= (17i - 3j - 10k) / | 17i - 3j - 10k |

= (17i - 3j - 10k) / √(17² + 3² + 10²)

= (17i - 3j - 10k) / √398