Let a→=αi^+j^−k^ and b→=2i^+j^−αk^,α>0 If the projection of a¯×b→ on the vector −i^+2j^−2k^ is 30, then α is equal to

a→×b→=1-αi^+α2−2j^+α-2k^ Projection of a→×b→ on −i^+2j^−2k^ =(a→×b→)⋅(−i^+2j^−2k^)3=30⇒2α2−α−91=0⇒2α2-14α+13α-91=0⇒2α+13α-7=0 ⇒α=7,−132

Let a→=αi^+j^−k^ and b→=2i^+j^−αk^,α>0 If the projection of a¯×b→ on the vector −i^+2j^−2k^ is 30, then α is equal to

a→×b→=1-αi^+α2−2j^+α-2k^ Projection of a→×b→ on −i^+2j^−2k^ =(a→×b→)⋅(−i^+2j^−2k^)3=30⇒2α2−α−91=0⇒2α2-14α+13α-91=0⇒2α+13α-7=0 ⇒α=7,−132

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Let $\vec{a} = α \hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = 2 \hat{i} + \hat{j} - α \hat{k}, α > 0$ If the projection of $\bar{a} \times \vec{b}$ on the vector $- \hat{i} + 2 \hat{j} - 2 \hat{k}$ is 30, then $α$ is equal to

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13/2

answer is D.

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Detailed Solution

$\vec{a} \times \vec{b} = (1 - α) \hat{i} + (α^{2} - 2) \hat{j} + (α - 2) \hat{k} Projection of \vec{a} \times \vec{b} on - \hat{i} + 2 \hat{j} - 2 \hat{k} \begin{array}{l} = \frac{(\vec{a} \times \vec{b}) \cdot (- \hat{i} + 2 \hat{j} - 2 \hat{k})}{3} = 30 \\ \Rightarrow 2 α^{2} - α - 91 = 0 \Rightarrow 2 α^{2} - 14 α + 13 α - 91 = 0 \\ \Rightarrow (2 α + 13) (α - 7) = 0 \end{array} \Rightarrow α = 7, - \frac{13}{2}$