Let a→=αi^+j^+βk^ and b→=3i^−5j^+4k^ be two vectors, such that a→×b→=−i^+9j^+12k^. Then the length of the projection of b→-2a→ on b→+a→ is equal to

Let a→=αi^+j^+βk^, b→=3i^−5j^+4k^a→×b→=−i^+9j^+12k^⇒i^j^k^α1β3−54⇒(4+5β)i^+(3β−4α)j^+(−5α−3)k^=−i^+9j^+12k^∴4+5β=−1,3β−4α=9,−5α−3=12β=−1, α=−3∴a→=−3i^+j^−k^,b→=3i^−5j^+4k^∴a→+b˙=−4j^+3k^|a→|2=11,|b→|2=50⇒−9+(−5)−4=−18∴ Projection of b→-2a→ on b→+a→ is(b→−2a→)⋅(a→+b→)|a→+b→|=|b→|2−2|a→|2−(a→⋅b→)|a→+b→|=50−22−(−18)5=465Ans. 465

Let a→=αi^+j^+βk^ and b→=3i^−5j^+4k^ be two vectors, such that a→×b→=−i^+9j^+12k^. Then the length of the projection of b→-2a→ on b→+a→ is equal to

Let a→=αi^+j^+βk^, b→=3i^−5j^+4k^a→×b→=−i^+9j^+12k^⇒i^j^k^α1β3−54⇒(4+5β)i^+(3β−4α)j^+(−5α−3)k^=−i^+9j^+12k^∴4+5β=−1,3β−4α=9,−5α−3=12β=−1, α=−3∴a→=−3i^+j^−k^,b→=3i^−5j^+4k^∴a→+b˙=−4j^+3k^|a→|2=11,|b→|2=50⇒−9+(−5)−4=−18∴ Projection of b→-2a→ on b→+a→ is(b→−2a→)⋅(a→+b→)|a→+b→|=|b→|2−2|a→|2−(a→⋅b→)|a→+b→|=50−22−(−18)5=465Ans. 465

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Let $\vec{a} = α \hat{i} + \hat{j} + β \hat{k} and \vec{b} = 3 \hat{i} - 5 \hat{j} + 4 \hat{k}$ be two vectors, such that $\vec{a} \times \vec{b} = - \hat{i} + 9 \hat{j} + 12 \hat{k}$ . Then the length of the projection of $\vec{b} - 2 \vec{a}$ on $\vec{b} + \vec{a}$ is equal to

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$\frac{46}{5}$

$\frac{39}{5}$

answer is D.

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Detailed Solution

$\begin{array}{l} Let \vec{a} = α \hat{i} + \hat{j} + β \hat{k}, \vec{b} = 3 \hat{i} - 5 \hat{j} + 4 \hat{k} \\ \vec{a} \times \vec{b} = - \hat{i} + 9 \hat{j} + 12 \hat{k} \\ \Rightarrow |\begin{array}{ccr} \hat{i} & \hat{j} & \hat{k} \\ α & 1 & β \\ 3 & - 5 & 4 \end{array}| \\ \Rightarrow (4 + 5 β) \hat{i} + (3 β - 4 α) \hat{j} + (- 5 α - 3) \hat{k} \\ = - \hat{i} + 9 \hat{j} + 12 \hat{k} \\ ∴ 4 + 5 β = - 1, 3 β - 4 α = 9, - 5 α - 3 = 12 \\ β = - 1, α = - 3 \\ ∴ \vec{a} = - 3 \hat{i} + \hat{j} - \hat{k}, \vec{b} = 3 \hat{i} - 5 \hat{j} + 4 \hat{k} \\ ∴ \vec{a} + \dot{b} = - 4 \hat{j} + 3 \hat{k} \\ | \vec{a} |^{2} = 11, | \vec{b} |^{2} = 50 \\ \Rightarrow - 9 + (- 5) - 4 = - 18 \end{array}$

$∴$ Projection of $\vec{b} - 2 \vec{a}$ on $\vec{b} + \vec{a}$ is

$\begin{array}{l} \frac{(\vec{b} - 2 \vec{a}) \cdot (\vec{a} + \vec{b})}{| \vec{a} + \vec{b} |} \\ = \frac{| \vec{b} |^{2} - 2 | \vec{a} |^{2} - (\vec{a} \cdot \vec{b})}{| \vec{a} + \vec{b} |} = \frac{50 - 22 - (- 18)}{5} = \frac{46}{5} \end{array}$