Let α =4 i+3 j+5 k and β = i+2 j-4 k. Let β _1 be parallel to α and β _2 be perpendicular to α. If β = β _1+ β _2, then the value of 5 β _2· ( i+ j+…

Given that $\vec{\alpha }=4\hat{i}+3\hat{j}+5\hat{k}$α=4i^+3j^​+5k^ and $\vec{\beta }=\hat{i}+2\hat{j}-4\hat{k}$β​=i^+2j^​−4k^ and $\vec{\beta }=\vec{{\beta }_{1}}+\vec{{\beta }_{2}}$β​=β1​​+β2​​

$$\Rightarrow \vec{{\beta }_{2}}=\vec{\beta }-\vec{{\beta }_{1}}..........(1)$$⇒β2​​=β​−β1​​..........(1)

since ${\vec{\beta }}_{1}$β​1​ is parallel to $\vec{\alpha }$α, $\vec{{\beta }_{1}}=t\vec{\alpha }$β1​​=tα

$$\Rightarrow \vec{{\beta }_{1}}=t(4\hat{i}+3\hat{j}+5\hat{k})=4t\hat{i}+3t\hat{j}+5t\hat{k}.........(2)$$⇒β1​​=t(4i^+3j^​+5k^)=4ti^+3tj^​+5tk^.........(2)

Substituting the values of $\vec{{\beta }_{1}}$β1​​ and $\vec{\alpha }$α in $(1)$(1), we get

$$\vec{{\beta }_{2}}=\hat{i}+2\hat{j}-4\hat{k}-(4t\hat{i}+3t\hat{j}+5t\hat{k})=(1-4t)\hat{i}+(2-3t)\hat{j}+(-4-5t)\hat{k}............(3)$$β2​​=i^+2j^​−4k^−(4ti^+3tj^​+5tk^)=(1−4t)i^+(2−3t)j^​+(−4−5t)k^............(3)

since $\vec{{\beta }_{2}}$β2​​ is perpendicular to $\vec{\alpha }$α, so $\vec{{\beta }_{2}}\cdot \vec{\alpha }=0$β2​​⋅α=0

$$\Rightarrow [(1-4t)\hat{i}+(2-3t)\hat{j}+(-4-5t)\hat{k}]\cdot [4\hat{i}+3\hat{j}+5\hat{k}]=0$$⇒[(1−4t)i^+(2−3t)j^​+(−4−5t)k^]⋅[4i^+3j^​+5k^]=0

$$\Rightarrow 4(1-4t)+3(2-3t)+5(-4-5t)=0$$⇒4(1−4t)+3(2−3t)+5(−4−5t)=0

$$\Rightarrow 4-16t+6-9t-20-25t=0$$⇒4−16t+6−9t−20−25t=0

$\Rightarrow -50t=10$⇒−50t=10$\Rightarrow t=\frac{-1}{5}$⇒t=5−1​

From $(2)$(2) and $(3)$(3), we get $\vec{{\beta }_{1}}=\frac{-1}{5}(4\hat{i}+3\hat{j}+5\hat{k})$β1​​=5−1​(4i^+3j^​+5k^)

$$\vec{{\beta }_{2}}=\frac{9}{5}\hat{i}+\frac{13}{5}\hat{j}-3\hat{k}=\frac{1}{5}(9\hat{i}+13\hat{j}-15\hat{k})$$β2​​=59​i^+513​j^​−3k^=51​(9i^+13j^​−15k^)

So, the value of $5{\vec{\beta }}_{2}\cdot (\hat{i}+\hat{j}+\hat{k})$5β​2​⋅(i^+j^​+k^) will be,

$$=5\times \frac{1}{5}(9\hat{i}+13\hat{j}-15\hat{k})\cdot (\hat{i}+\hat{j}+\hat{k})=9+13-15=7$$=5×51​(9i^+13j^​−15k^)⋅(i^+j^​+k^)=9+13−15=7