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

β1→=k1α→,β2→=β1→-β=k1α→-2i^−j^+3k^ β2→=k13i^+j^-2i^+j^-3k^=3k1-2i^+k1+1j-3k^ α¯.β2¯=03k1−2.3+k1+1.1=09k1−6+k1+1=0k1=12β1¯=123i¯+j¯,β2¯=-12i^+32j^-3k^ ⇒β1¯×β2¯=i^j^k^3/21/20-1/23/2-3 ∴ 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

β1→=k1α→,β2→=β1→-β=k1α→-2i^−j^+3k^ β2→=k13i^+j^-2i^+j^-3k^=3k1-2i^+k1+1j-3k^ α¯.β2¯=03k1−2.3+k1+1.1=09k1−6+k1+1=0k1=12β1¯=123i¯+j¯,β2¯=-12i^+32j^-3k^ ⇒β1¯×β2¯=i^j^k^3/21/20-1/23/2-3 ∴ β→1×β→2=12(−3i^+9j^+5k^)

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

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

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

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

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

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Detailed Solution

$\vec{β_{1}} = k_{1} \vec{α}, \vec{β_{2}} = \vec{β_{1}} - β = k_{1} \vec{α} - (2 \hat{i} - \hat{j} + 3 \hat{k}) \vec{β_{2}} = k_{1} (3 \hat{i} + \hat{j}) - 2 \hat{i} + \hat{j} - 3 \hat{k} = (3 k_{1} - 2) \hat{i} + (k_{1} + 1) j - 3 \hat{k} \begin{array}{l} \bar{α} . \bar{β_{2}} = 0 \\ (3 k_{1} - 2) . 3 + (k_{1} + 1) . 1 = 0 \\ 9 k_{1} - 6 + k_{1} + 1 = 0 \\ k_{1} = \frac{1}{2} \end{array}$

$\bar{β_{1}} = \frac{1}{2} (3 \bar{i} + \bar{j}), \bar{β_{2}} = \frac{- 1}{2} \hat{i} + \frac{3}{2} \hat{j} - 3 \hat{k} \Rightarrow \bar{β_{1}} \times \bar{β_{2}} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 3 / 2 & 1 / 2 & 0 \\ - 1 / 2 & 3 / 2 & - 3 \end{matrix}| ∴ {\vec{β}}_{1} \times {\vec{β}}_{2} = \frac{1}{2} (- 3 \hat{i} + 9 \hat{j} + 5 \hat{k})$