Let a→=3i^−j^+2k^,b→=a→×(i^−2k^) and c→=b→×k^. Then the projection of c→−2j^ on a→ is :

b→=a→×(i^−3k^)=i^j^k^3−1210−2=2i^+8j^+k^c→=b→×k^=8i^−2j^c→−2j^=8i^−4j^Projection of (i^−2j^) on a→(c→−2j^)⋅a^=⟨8,−4,0⟩⋅⟨3,−1,2⟩14=2814=214

Let a→=3i^−j^+2k^,b→=a→×(i^−2k^) and c→=b→×k^. Then the projection of c→−2j^ on a→ is :

b→=a→×(i^−3k^)=i^j^k^3−1210−2=2i^+8j^+k^c→=b→×k^=8i^−2j^c→−2j^=8i^−4j^Projection of (i^−2j^) on a→(c→−2j^)⋅a^=⟨8,−4,0⟩⋅⟨3,−1,2⟩14=2814=214

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

Let $\vec{a} = 3 \hat{i} - \hat{j} + 2 \hat{k}, \vec{b} = \vec{a} \times (\hat{i} - 2 \hat{k}) and \vec{c} = \vec{b} \times \hat{k} .$ Then the projection of $\vec{c} - 2 \hat{j} on \vec{a}$ is :

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

$2 \sqrt{7}$

$3 \sqrt{7}$

$2 \sqrt{14}$

$\sqrt{14}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\vec{b} = \vec{a} \times (\hat{i} - 3 \hat{k})$
$\begin{array}{l} = |\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & - 1 & 2 \\ 1 & 0 & - 2 \end{array}| = 2 \hat{i} + 8 \hat{j} + \hat{k} \\ \vec{c} = \vec{b} \times \hat{k} = 8 \hat{i} - 2 \hat{j} \\ \vec{c} - 2 \hat{j} = 8 \hat{i} - 4 \hat{j} \end{array}$
Projection of $(\hat{i} - 2 \hat{j}) on \vec{a}$
$\begin{array}{l} (\vec{c} - 2 \hat{j}) \cdot \hat{a} = \frac{⟨ 8, - 4, 0 ⟩ \cdot ⟨ 3, - 1, 2 ⟩}{\sqrt{14}} \\ = \frac{28}{\sqrt{14}} = 2 \sqrt{14} \end{array}$