If D=7i-3j+2k and E=4i+5j-3k. Deduce angle between D and E
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D=7i-3j+3k,E=6i+2j-2k
82.83
To deduce the angle between vectors D and E, we can use the dot product formula. The dot product of two vectors A and B is given by the formula:
A · B = |A| |B| cos(θ),
where |A| represents the magnitude (or length) of vector A, |B| represents the magnitude of vector B, and θ is the angle between the two vectors.
In this case, given vectors D and E, we can calculate their dot product as follows:
D · E = (7i - 3j + 2k) · (4i + 5j - 3k)
= (7 * 4) + (-3 * 5) + (2 * -3)
= 28 - 15 - 6
= 7.
Next, we need to find the magnitudes of D and E to substitute into the dot product formula. The magnitude of a vector can be found using the Pythagorean theorem:
|D| = sqrt((7^2) + (-3^2) + (2^2)),
= sqrt(49 + 9 + 4),
= sqrt(62).
|E| = sqrt((4^2) + (5^2) + (-3^2)),
= sqrt(16 + 25 + 9),
= sqrt(50).
Now we can substitute the magnitudes and the dot product into the dot product formula to find the angle θ:
7 = sqrt(62) * sqrt(50) * cos(θ).
Dividing both sides of the equation by sqrt(62) * sqrt(50), we obtain:
cos(θ) = 7 / (sqrt(62) * sqrt(50)).
Finally, we can solve for θ by taking the inverse cosine (arccos) of both sides:
θ = arccos(7 / (sqrt(62) * sqrt(50))).
Therefore, the angle between vectors D and E is given by θ = arccos(7 / (sqrt(62) * sqrt(50))).