Let $\overrightarrow{\mathrm{\alpha }}=4\widehat{\mathrm{i}}+3\widehat{\mathrm{j}}+5\widehat{\mathrm{k}}\text{ and }\overrightarrow{\beta }=\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}-4\widehat{\mathrm{k} }\text{.}$$\mathrm{Let} \overrightarrow{{\mathrm{\beta }}_1}$ be parallel to $\stackrel{\to }{\alpha }\text{and }\overrightarrow{{\beta }_2}$ be perpendicular to $\overrightarrow{\alpha }$ . If $\overrightarrow{\beta }={\overrightarrow{\mathrm{\beta }}}_1+{\overrightarrow{\mathrm{\beta }}}_2$ , then the value of $5{\overrightarrow{\mathrm{\beta }}}_2.(\widehat{\mathrm{i}}+\widehat{\mathrm{j}}+\widehat{\mathrm{k}})$ is 

Complete Solution:

Since β₁ is parallel to α, we can write it as:

β₁ = λα

where λ is a scalar. To find λ, we use the projection of β onto α:

β₁ = (β ⋅ α / |α|²) α

Calculate β ⋅ α:

β ⋅ α = (1)(4) + (2)(3) + (-4)(5) = 4 + 6 - 20 = -10

Calculate |α|²:

|α|² = 4² + 3² + 5² = 16 + 9 + 25 = 50

Substitute values for β₁:

β₁ = (-10 / 50) α = -1/5 α

Using α = 4i + 3j + 5k:

β₁ = -4/5 i - 3/5 j - k

Since β = β₁ + β₂, we can write:

β₂ = β - β₁

Substitute β = i + 2j - 4k and β₁ = -4/5 i - 3/5 j - k:

β₂ = (i + 2j - 4k) - (-4/5 i - 3/5 j - k)

= 9/5 i + 13/5 j - 3k

Multiply β₂ by 5:

5β₂ = 9i + 13j - 15k

Dot product with (i + j + k):

5β₂ ⋅ (i + j + k) = (9)(1) + (13)(1) + (-15)(1)

= 9 + 13 - 15 = 7

Final Answer:

5β₂ ⋅ (i + j + k) = 7