If a→ and b→ are non-zero non-collinear vectors, then [a→b→i^]i^+[a→b→j^]j^+[a→b→k^]k^ is equal to

Let a→×b→=xi^+yj^+zk^. Therefore, [a→b→i^]=(a→×b→)⋅i^=x[a→b→j^]=(a→×b→)⋅j^=y[a→b→k→]=(a→×b→)⋅k^=z Hence, [a→b→i^]i^+[a→b→j^]j^=[a→b→k^]k^=xi^+yj^+zk^=a→×b→hence [a→b→i^]i^+[a→b→j^]j^=[a→b→k^]k^=xi^+yj^+zk^=a→×b→

If a→ and b→ are non-zero non-collinear vectors, then [a→b→i^]i^+[a→b→j^]j^+[a→b→k^]k^ is equal to

Let a→×b→=xi^+yj^+zk^. Therefore, [a→b→i^]=(a→×b→)⋅i^=x[a→b→j^]=(a→×b→)⋅j^=y[a→b→k→]=(a→×b→)⋅k^=z Hence, [a→b→i^]i^+[a→b→j^]j^=[a→b→k^]k^=xi^+yj^+zk^=a→×b→hence [a→b→i^]i^+[a→b→j^]j^=[a→b→k^]k^=xi^+yj^+zk^=a→×b→

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

If $\vec{a} and \vec{b}$ are non-zero non-collinear vectors, then $[\vec{a} \vec{b} \hat{i}] \hat{i} + [\vec{a} \vec{b} \hat{j}] \hat{j} + [\vec{a} \vec{b} \hat{k}] \hat{k}$ 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

$\vec{a} + \vec{b}$

$\vec{a} \times \vec{b}$

$\vec{a} - \vec{b}$

$\vec{b} - \vec{a}$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} Let \vec{a} \times \vec{b} = x \hat{i} + y \hat{j} + z \hat{k} . Therefore, \\ [\vec{a} \vec{b} \hat{i}] = (\vec{a} \times \vec{b}) \cdot \hat{i} = x \\ [\vec{a} \vec{b} \hat{j}] = (\vec{a} \times \vec{b}) \cdot \hat{j} = y \\ [\vec{a} \vec{b} \vec{k}] = (\vec{a} \times \vec{b}) \cdot \hat{k} = z \\ Hence, [\vec{a} \vec{b} \hat{i}] \hat{i} + [\vec{a} \vec{b} \hat{j}] \hat{j} \\ = [\vec{a} \vec{b} \hat{k}] \hat{k} = x \hat{i} + y \hat{j} + z \hat{k} = \vec{a} \times \vec{b} \end{array}$

$\begin{array}{l} hence [\vec{a} \vec{b} \hat{i}] \hat{i} + [\vec{a} \vec{b} \hat{j}] \hat{j} \\ = [\vec{a} \vec{b} \hat{k}] \hat{k} = x \hat{i} + y \hat{j} + z \hat{k} = \vec{a} \times \vec{b} \end{array}$