Let α→ =3i^+j^ and β→ =2i^-j^+3k^. lf where β1→ is parallel to α→ β→2 is perpendicular to β →then β1→ ×β→2 is equal to :

β¯1=P r is vector β¯ on α¯=β¯·α¯α¯2α¯ =6-110(3i+j)=3i+j2β¯2=β¯-β¯1=2i-j+3k-3i+j2 =i-3j+6k2β¯1×β¯2=14ijk3101-36=14(i(6)-j(18)+k(-10)) =3i-9j-5k2

Let α→ =3i^+j^ and β→ =2i^-j^+3k^. lf where β1→ is parallel to α→ β→2 is perpendicular to β →then β1→ ×β→2 is equal to :

β¯1=P r is vector β¯ on α¯=β¯·α¯α¯2α¯ =6-110(3i+j)=3i+j2β¯2=β¯-β¯1=2i-j+3k-3i+j2 =i-3j+6k2β¯1×β¯2=14ijk3101-36=14(i(6)-j(18)+k(-10)) =3i-9j-5k2

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

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

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

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

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

answer is C.

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Detailed Solution

$\begin{array}{rcl} {\bar{β}}_{1} & = & P r is vector \bar{β} on \bar{α} = (\frac{\bar{β} \cdot \bar{α}}{{|\bar{α}|}^{2}}) \bar{α} \\ = & (\frac{6 - 1}{10}) (3 i + j) = \frac{3 i + j}{2} \\ {\bar{β}}_{2} & = & \bar{β} - {\bar{β}}_{1} = 2 i - j + 3 k - (\frac{3 i + j}{2}) \\ = & \frac{i - 3 j + 6 k}{2} \\ {\bar{β}}_{1} \times {\bar{β}}_{2} & = & \frac{1}{4} |\begin{matrix} i & j & k \\ 3 & 1 & 0 \\ 1 & - 3 & 6 \end{matrix}| = \frac{1}{4} (i (6) - j (18) + k (- 10)) \\ = & \frac{3 i - 9 j - 5 k}{2} \end{array}$