Let α→=4i^+3j^+5k^ and β→=i^+2j^−4k^. Let β1→ be parallel to α→ and β2→ be perpendicular to α→. If β→=β→1+β→2 , then the value of 5β→2· (i^+j^+k^) is

Given β→1 parallel to α→ ⇒ β→1=tα→β→2 perpendicular to α→ ⇒ β→2⋅α→=0 β→=β→1+β→2Now do dot product with α→ on both sides α→⋅β→=β→1⋅α→+β→2α→ (β→2⋅α→=0)(4i^+3j^+5k^)(i^+2j^−4k^)=(tα→)⋅α→+0 (β→1=tα→) 4+6−20=t(16+9+25)+0 −10=t(50)⇒t=−15β→1=−15(4i^+3j^+5k^)From β→=β→1+β→2β→2=β→−β→1=(i+2j−4k)−(−15(4i+3j+5k))β→2=15(9i^+13j^−15k^) 5β→2⋅(i^+j^+k^)=9+13−15 5β→2⋅(i^+j^+k^)=7

Let α→=4i^+3j^+5k^ and β→=i^+2j^−4k^. Let β1→ be parallel to α→ and β2→ be perpendicular to α→. If β→=β→1+β→2 , then the value of 5β→2· (i^+j^+k^) is

Given β→1 parallel to α→ ⇒ β→1=tα→β→2 perpendicular to α→ ⇒ β→2⋅α→=0 β→=β→1+β→2Now do dot product with α→ on both sides α→⋅β→=β→1⋅α→+β→2α→ (β→2⋅α→=0)(4i^+3j^+5k^)(i^+2j^−4k^)=(tα→)⋅α→+0 (β→1=tα→) 4+6−20=t(16+9+25)+0 −10=t(50)⇒t=−15β→1=−15(4i^+3j^+5k^)From β→=β→1+β→2β→2=β→−β→1=(i+2j−4k)−(−15(4i+3j+5k))β→2=15(9i^+13j^−15k^) 5β→2⋅(i^+j^+k^)=9+13−15 5β→2⋅(i^+j^+k^)=7

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Let $\vec{α} = 4 \hat{i} + 3 \hat{j} + 5 \hat{k}$ and $\vec{β} = \hat{i} + 2 \hat{j} - 4 \hat{k}$ . Let $\vec{β_{1}}$ be parallel to $\vec{α}$ and $\vec{β_{2}}$ be perpendicular to $\vec{α}$ . If $\vec{β} = {\vec{β}}_{1} + {\vec{β}}_{2}$ , then the value of $5 {\vec{β}}_{2} \cdot (\hat{i} + \hat{j} + \hat{k})$ is

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Detailed Solution

Given ${\vec{β}}_{1}$ parallel to $\vec{α} \Rightarrow {\vec{β}}_{1} = t \vec{α}$
${\vec{β}}_{2}$ perpendicular to $\vec{α} \Rightarrow {\vec{β}}_{2} \cdot \vec{α} = 0$
$\vec{β} = {\vec{β}}_{1} + {\vec{β}}_{2}$
Now do dot product with $\vec{α}$ on both sides
$\vec{α} \cdot \vec{β} = {\vec{β}}_{1} \cdot \vec{α} + {\vec{β}}_{2} \vec{α} ({\vec{β}}_{2} \cdot \vec{α} = 0)$

$(4 \hat{i} + 3 \hat{j} + 5 \hat{k}) (\hat{i} + 2 \hat{j} - 4 \hat{k}) = (t \vec{α}) \cdot \vec{α} + 0 ({\vec{β}}_{1} = t \vec{α}) 4 + 6 - 20 = t (16 + 9 + 25) + 0 - 10 = t (50) \Rightarrow t = \frac{- 1}{5}$

${\vec{β}}_{1} = \frac{- 1}{5} (4 \hat{i} + 3 \hat{j} + 5 \hat{k})$

From $\vec{β} = {\vec{β}}_{1} + {\vec{β}}_{2}$

${\vec{β}}_{2} = \vec{β} - {\vec{β}}_{1} = (i + 2 j - 4 k) - (\frac{- 1}{5} (4 i + 3 j + 5 k))$

${\vec{β}}_{2} = \frac{1}{5} (9 \hat{i} + 13 \hat{j} - 15 \hat{k}) 5 {\vec{β}}_{2} \cdot (\hat{i} + \hat{j} + \hat{k}) = 9 + 13 - 15 5 {\vec{β}}_{2} \cdot (\hat{i} + \hat{j} + \hat{k}) = 7$