Let α¯=3i¯+j¯ and β¯=2i¯−j¯+3k¯. If β¯=β¯1−β¯2 where β¯1 is parallel to α¯ and β¯2 is perpendicular to α¯ then β¯1×β¯2 is equal to

∵β¯1=λα¯⇒β¯1=λ3i¯+j¯∴ β¯=β¯1−β¯2 β¯2=β¯1−β¯⇒β¯2=3λ−2i¯+λ+1j¯−3k¯∴β¯2.α¯=0⇒λ=12∴β¯1×β¯2=12−3i¯+9j¯+5k¯

Let α¯=3i¯+j¯ and β¯=2i¯−j¯+3k¯. If β¯=β¯1−β¯2 where β¯1 is parallel to α¯ and β¯2 is perpendicular to α¯ then β¯1×β¯2 is equal to

∵β¯1=λα¯⇒β¯1=λ3i¯+j¯∴ β¯=β¯1−β¯2 β¯2=β¯1−β¯⇒β¯2=3λ−2i¯+λ+1j¯−3k¯∴β¯2.α¯=0⇒λ=12∴β¯1×β¯2=12−3i¯+9j¯+5k¯

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Let $\bar{α} = 3 \bar{i} + \bar{j}$ and $\bar{β} = 2 \bar{i} - \bar{j} + 3 \bar{k}$ . If $\bar{β} = {\bar{β}}_{1} - {\bar{β}}_{2}$ where ${\bar{β}}_{1}$ is parallel to $\bar{α}$ and ${\bar{β}}_{2}$ is perpendicular to $\bar{α}$ then ${\bar{β}}_{1} \times {\bar{β}}_{2}$ is equal to

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$- 3 \bar{i} - 9 \bar{j} + 5 \bar{k}$

$\frac{1}{2} (- 3 \bar{i} + 9 \bar{j} + 5 \bar{k})$

$\frac{1}{2} (3 \bar{i} - 9 \bar{j} + 5 \bar{k})$

$3 \bar{i} - 9 \bar{j} - 5 \bar{k}$

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$∵ {\bar{β}}_{1} = λ \bar{α}$

$\Rightarrow {\bar{β}}_{1} = λ (3 \bar{i} + \bar{j})$

$∴ \bar{β} = {\bar{β}}_{1} - {\bar{β}}_{2}$

${\bar{β}}_{2} = {\bar{β}}_{1} - \bar{β}$

$\Rightarrow {\bar{β}}_{2} = (3 λ - 2) \bar{i} + (λ + 1) \bar{j} - 3 \bar{k}$

$∴ {\bar{β}}_{2} . \bar{α} = 0$

$\Rightarrow λ = \frac{1}{2}$

$∴ {\bar{β}}_{1} \times {\bar{β}}_{2} = \frac{1}{2} (- 3 \bar{i} + 9 \bar{j} + 5 \bar{k})$

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Let α¯=3i¯+j¯ and β¯=2i¯−j¯+3k¯. If β¯=β¯1−β¯2 where β¯1 is parallel to α¯ and β¯2 is perpendicular to α¯ then β¯1×β¯2 is equal to

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Let $\bar{α} = 3 \bar{i} + \bar{j}$ and $\bar{β} = 2 \bar{i} - \bar{j} + 3 \bar{k}$ . If $\bar{β} = {\bar{β}}_{1} - {\bar{β}}_{2}$ where ${\bar{β}}_{1}$ is parallel to $\bar{α}$ and ${\bar{β}}_{2}$ is perpendicular to $\bar{α}$ then ${\bar{β}}_{1} \times {\bar{β}}_{2}$ is equal to