Resolved part of vector a→ and along vector b→ is a1→ and at perpendicular to b→ is a→2, then a→1×a→2 is equal to

a→1=(a→⋅b^)b^=(a→⋅b→)b→|b→|2⇒a→2=a→−a→1=a→−(a→⋅b→)b→|b→|2 Thus, a→1×a→2=(a→⋅b→)b→|b→|2×a→−(a→⋅b→)b→|b→|2=(a→⋅b→)(b→×a→)|b→|2

Resolved part of vector a→ and along vector b→ is a1→ and at perpendicular to b→ is a→2, then a→1×a→2 is equal to

a→1=(a→⋅b^)b^=(a→⋅b→)b→|b→|2⇒a→2=a→−a→1=a→−(a→⋅b→)b→|b→|2 Thus, a→1×a→2=(a→⋅b→)b→|b→|2×a→−(a→⋅b→)b→|b→|2=(a→⋅b→)(b→×a→)|b→|2

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

Resolved part of vector $\vec{a}$ and along vector $\vec{b}$ is $\vec{a_{1}}$ and at perpendicular to $\vec{b} is {\vec{a}}_{2}, then {\vec{a}}_{1} \times {\vec{a}}_{2}$ 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

$\frac{(\vec{a} \cdot \vec{b}) (\vec{b} \times \vec{a})}{| \vec{b} |^{2}}$

$\frac{(\vec{a} \cdot \vec{b}) \vec{a}}{| \vec{a} |^{2}}$

$\frac{(\vec{a} \cdot \vec{b}) (\vec{b} \times \vec{a})}{| \vec{b} \times \vec{a} |}$

$\frac{(\vec{a} \times \vec{b}) \cdot \vec{b}}{| \vec{b} |^{2}}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} {\vec{a}}_{1} = (\vec{a} \cdot \hat{b}) \hat{b} = \frac{(\vec{a} \cdot \vec{b}) \vec{b}}{| \vec{b} |^{2}} \\ \Rightarrow {\vec{a}}_{2} = \vec{a} - {\vec{a}}_{1} = \vec{a} - \frac{(\vec{a} \cdot \vec{b}) \vec{b}}{| \vec{b} |^{2}} \\ Thus, {\vec{a}}_{1} \times {\vec{a}}_{2} = \frac{(\vec{a} \cdot \vec{b}) \vec{b}}{| \vec{b} |^{2}} \times (\vec{a} - \frac{(\vec{a} \cdot \vec{b}) \vec{b}}{| \vec{b} |^{2}}) \\ = \frac{(\vec{a} \cdot \vec{b}) (\vec{b} \times \vec{a})}{| \vec{b} |^{2}} \end{array}$