Given that a→,b→,p→,q→ are four vectors such that a→+b→=μp→,b→⋅q→=0 and (b→)2=1 where μ is a scalar. Then |(a→⋅q→)p→−(p→⋅q→)a→|is equal to

a→+b→ =μp→ b→⋅q→=0,|b→|2=1∵ a→+b→=μp→⇒ (a→+b→)×a→=μp→×a→,b→×a→=μp→×a→⇒ q→×(b→×a→)=μq→×(p→×a→)⇒ (q→⋅a→)b→−(q→⋅b→)a→=μq→×(p→×a→)⇒ (q→⋅a→)b→=μq→×(p→×a→)∵ a→+b→=μp→⇒ q→⋅(a→+b→)=μq→⋅p→⇒ q→⋅a→+q→⋅b→=μp→⋅q→⇒ μ=q→⋅a→p→⋅q→⇒ (q→⋅a→)b→=q→⋅a→p→⋅q→[(q→⋅a→)⋅p→−(q→⋅p→)a→]⇒ |(q→⋅a→)p→−(q→⋅p→)a→| =|(p→⋅q→)b→|=|(p→⋅q→)|⋅|b→|⇒ |(q→⋅a→)p→−(q→⋅p→)a→|=|p→⋅q→|

Given that a→,b→,p→,q→ are four vectors such that a→+b→=μp→,b→⋅q→=0 and (b→)2=1 where μ is a scalar. Then |(a→⋅q→)p→−(p→⋅q→)a→|is equal to

a→+b→ =μp→ b→⋅q→=0,|b→|2=1∵ a→+b→=μp→⇒ (a→+b→)×a→=μp→×a→,b→×a→=μp→×a→⇒ q→×(b→×a→)=μq→×(p→×a→)⇒ (q→⋅a→)b→−(q→⋅b→)a→=μq→×(p→×a→)⇒ (q→⋅a→)b→=μq→×(p→×a→)∵ a→+b→=μp→⇒ q→⋅(a→+b→)=μq→⋅p→⇒ q→⋅a→+q→⋅b→=μp→⋅q→⇒ μ=q→⋅a→p→⋅q→⇒ (q→⋅a→)b→=q→⋅a→p→⋅q→[(q→⋅a→)⋅p→−(q→⋅p→)a→]⇒ |(q→⋅a→)p→−(q→⋅p→)a→| =|(p→⋅q→)b→|=|(p→⋅q→)|⋅|b→|⇒ |(q→⋅a→)p→−(q→⋅p→)a→|=|p→⋅q→|

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Given that $\vec{a}, \vec{b}, \vec{p}, \vec{q}$ are four vectors such that $\vec{a} + \vec{b} = μ \vec{p}, \vec{b} \cdot \vec{q} = 0 and (\vec{b})^{2} = 1$ where $μ$ is a scalar. Then $| (\vec{a} \cdot \vec{q}) \vec{p} - (\vec{p} \cdot \vec{q}) \vec{a} |$ is equal to

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$2 | \vec{p} \cdot \vec{q} |$

$| \vec{p} \times \vec{q} |$

$(1 / 2) | \vec{p} \cdot \vec{q} |$

$| \vec{p} . \vec{q} |$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} \vec{a} + \vec{b} = μ \vec{p} \vec{b} \cdot \vec{q} = 0, | \vec{b} |^{2} = 1 \\ ∵ \vec{a} + \vec{b} = μ \vec{p} \\ \Rightarrow (\vec{a} + \vec{b}) \times \vec{a} = μ \vec{p} \times \vec{a}, \vec{b} \times \vec{a} = μ \vec{p} \times \vec{a} \\ \Rightarrow \vec{q} \times (\vec{b} \times \vec{a}) = μ \vec{q} \times (\vec{p} \times \vec{a}) \\ \Rightarrow (\vec{q} \cdot \vec{a}) \vec{b} - (\vec{q} \cdot \vec{b}) \vec{a} = μ \vec{q} \times (\vec{p} \times \vec{a}) \\ \Rightarrow (\vec{q} \cdot \vec{a}) \vec{b} = μ \vec{q} \times (\vec{p} \times \vec{a}) \\ ∵ \vec{a} + \vec{b} = μ \vec{p} \\ \Rightarrow \vec{q} \cdot (\vec{a} + \vec{b}) = μ \vec{q} \cdot \vec{p} \\ \Rightarrow \vec{q} \cdot \vec{a} + \vec{q} \cdot \vec{b} = μ \vec{p} \cdot \vec{q} \\ \Rightarrow μ = \frac{\vec{q} \cdot \vec{a}}{\vec{p} \cdot \vec{q}} \\ \Rightarrow (\vec{q} \cdot \vec{a}) \vec{b} = \frac{\vec{q} \cdot \vec{a}}{\vec{p} \cdot \vec{q}} [(\vec{q} \cdot \vec{a}) \cdot \vec{p} - (\vec{q} \cdot \vec{p}) \vec{a}] \\ \Rightarrow | (\vec{q} \cdot \vec{a}) \vec{p} - (\vec{q} \cdot \vec{p}) \vec{a} | \\ = | (\vec{p} \cdot \vec{q}) \vec{b} | = | (\vec{p} \cdot \vec{q}) | \cdot | \vec{b} | \\ \Rightarrow | (\vec{q} \cdot \vec{a}) \vec{p} - (\vec{q} \cdot \vec{p}) \vec{a} | = | \vec{p} \cdot \vec{q} | \end{array}$