Respuesta :
Explanation:
Given that:
a→ = (i^ - 2j^ + 2k^)/3
b→ = (-4i^ - 3k^)/5
c→ = j^
Now,
| a | = √{(1/9) + (4/9) + (4/9)} = √{(1+4+4)/9} = √{(9/9)} = 1
| b | = √{(9/25) + (16/25)} = √{(9 + 16)/25} = √(25/25) = 1
| c | = 1
Now,
Let us assume that
⇛d→ = xi^ + yj^ + zk^ which makes equal angle
⇛α, β, γ with a→, b→, c→ respectively.
So, According to the statement,
⇛cos α = cos β = cos γ
⇛{(d→.a→)/(| d→ || a→ |)} = {(d→.b→)/(| d→ || c→|)} = {(d→.c→)/(| d→ || c→ |)}
⇛[{x-2y+2z}/{3√(x²+y²+z²)} = [{-4x - 3z}/{5√(x²+y²+z²)} = {y/√(x²+y²+z²)}
⇛{(x² - 2y + 2z)/3} = {(-4x - 3z)/5} = y
On Taking first and third member, we get
⇛{(x² - 2y + 2z)/3} = y
⇛{(x² - 2y + 2z)/3} = y/1
On applying cross multiplication then
⇛1(x² - 2y + 2z) = 3(y)
⇛x² - 2y + 2z = 3y
⇛x² - 2y + 2z - 3y = 0
⇛x² - 2y - 3y + 2z = 0
⇛x² - 5y + 2z = 0 --------Eqn(1)
Now,
Taking second and third member, we get
⇛ {(-4x - 3z)/5} = y
⇛ {(-4x - 3z)/5} = y/1
On applying cross multiplication then
⇛1(-4x - 3z) = 5(y)
⇛-4x - 3z = 5y
⇛-4x - 5y - 3z = 0 --------Eqn(2)
Now, solving equation (1) and equation (2) using cross multiplication method, we get
⇛{x/(-15 - 10)} = {y/(8 - 3)} = {z/(5 + 20)}
⇛(x/-25) = (y/5) = (z/25)
⇛(x/-5) = (y/1) = (z/5)
⇛(x/5) = (y/-1) = (z/-5)
⇛x = 5k
⇛y = -k
⇛z = -5k
So,
⇛d→ = 5k^ - kj^ - 5kk^
As,
⇛| d→ | = 1
⇛√(25k² + k²+ 25k²) = √(51)
⇛√(26k² + 25k²) = √(51)
⇛51k² = √(51)
⇛k² = 1
⇛k = ±1
Thus,
⇛d→ = ± (5i^ - j^ - 5k^)
Answer: So, required vector is
⇛d→ = 5i^ - j^ - 5k^
and
⇛| d→ | = √(25 + 1 + 25) = √(26 + 25) = √(51)
Please let me know if you have any other questions.
It should be noted that the vectors of magnitude √51 that makes equation angles with the three vectors will be 5i - j - 5k.
How to calculate the vector
From the information given, the following can be depicted:
a = (i^ - 2j^ + 2k^)/3
b = (-4i^ - 3k^)/5
c = j^
After applying cross multiplication, the first equation will be x² - 5y + 2z = 0 and the second equation will be -4x - 5y - 3z.
Solving the equations further will be:
[x/(-15 - 10)] = [(y/(8 - 3)] = [z(5 + 20)]
= (x/-25) = (y/1) = (z/5)
= (x/5) = (y/-1) = (z/-5)
Therefore, x = 5k, y = -k, and z = -5k.
In conclusion, the required vector is 5i - j - 5k.
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