Which one of the following vectors of magnitude √51 makes equation angles with three vectors a= (i^ - 2j^ + 2k^)/3, b= (-4i^ - 3k^)/5 and = c =j^ ?

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.

do you have a formula possibly