the vector parallel to 4i - 3j + 5k and whose lengths is the arithmetic mean of lengths of two vectors 2i - 4j + 4k and i+ 6j + 3k is
Answers
Given:
The vector parallel to 4i - 3j + 5k and lengths is the arithmetic mean of lengths of two vectors 2i - 4j + 4k and i+ 6j + 3k.
To find:
The parallel vector.
Solution:
From given, we have,
The vector parallel to 4i - 3j + 5k
⇒ The parallel vector is given by,
= (4i - 3j + 5k) × 1/√(4² + 3² + 5²)
= (4i - 3j + 5k)/√50
The lengths is the arithmetic mean of lengths of two vectors 2i - 4j + 4k and i+ 6j + 3k.
The magnitude of 2i - 4j + 4k is
= √(2² + 4² + 4²)
= 6
The magnitude of i +6j + 3k is
= √(1² + 6² + 3²)
= √46
⇒ The arithmetic mean of lengths
(6 + √46)/2
So, the required vector is,
= (4i - 3j + 5k)/√50 × (6 + √46)/2
= (3√2 + √23)/10 (4i - 3j + 5k)
Answer:
(4i-3j+5k) /√2
Explanation:
k=|b|+|c|/2
=√36+√16/2
=6+4/2
=10/2
k=5
D=k(a) /|a|
=5(4i-3j+5k)/√50
=5(4i-3j+5k)/5√2
=4i-3j+5k/√2
in question there is a mistake that is I+√6j+3k not I+6j+3k
thank you❤