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QJEEM23Ma9c66: JEE Main 2023 Physics - Class 11 JEEM23Ma9c66 Solution: Let and . If is a vector such that and , then is equal to…

Let a⃗=i^+2j^+3k^\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k} and b⃗=i^+j^−k^\vec{b}=\hat{i}+\hat{j}-\hat{k} . If c⃗\vec{c} is a vector such that a⃗⋅c⃗=11,b⃗⋅(a⃗×c⃗)=27\vec{a} \cdot \vec{c}=11,\vec{b} \cdot(\vec{a} \times \vec{c})=27 and b⃗⋅c⃗=−3∣b⃗∣\vec{b} \cdot \vec{c}=-\sqrt{3}|\vec{b}| , then ∣a⃗×c⃗∣2|\vec{a} \times \vec{c}|^{2} is equal to _________.

Step-by-Step Verified Solution

Given,

amp;a⃗=i^+2j^+3k^amp;b⃗=i^+j^−k^amp;a⃗⋅c⃗=11amp;b⃗⋅(a⃗×c⃗)=27amp;b⃗⋅c⃗=−3∣b⃗∣amp;(b⃗×a⃗)⋅c⃗=27\begin{aligned}& \vec{a}=\hat{i}+2 \hat{j}+3 \hat{k} \\\\& \vec{b}=\hat{i}+\hat{j}-\hat{k} \\\\& \vec{a} \cdot \vec{c}=11 \\\\& \vec{b} \cdot(\vec{a} \times \vec{c})=27 \\\\& \vec{b} \cdot \vec{c}=-\sqrt{3}|\vec{b}| \\\\& (\vec{b} \times \vec{a}) \cdot \vec{c}=27\end{aligned}

 Let c⃗=c1i^+c2j^+c3k^\text { Let } \vec{c}=c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k}

As a⃗⋅c⃗=11\vec{a} \cdot \vec{c}=11

∴\therefore c1+2c2+3c3=11c_1+2 c_2+3 c_3=11 ......(i)

Also, b⃗⋅c⃗=−3∣b⃗∣\vec{b} \cdot \vec{c}=-\sqrt{3}|\vec{b}|

amp;∴c1+c2−c3=−33amp;⇒c1+c2−c3=−3......(ii)\begin{aligned}& \therefore c_1+c_2-c_3=-\sqrt{3} \sqrt{3} \\\\& \Rightarrow c_1+c_2-c_3=-3 ......(ii)\end{aligned}

Also, b⃗⋅(a⃗×c⃗)=27\vec{b} \cdot(\vec{a} \times \vec{c})=27

∴\therefore 5c1−4c2+c3=275 c_1-4 c_2+c_3=27 ...........(iii)

From (i), (ii) & (iii)

c⃗=3i^−2j^+4k^\vec{c}=3 \hat{i}-2 \hat{j}+4 \hat{k}

amp;∣a⃗×c⃗∣2=∣i^amp;j^amp;k^1amp;2amp;33amp;−2amp;+4∣2amp;=∣14i^+5j^−8k^∣2amp;=142+52+82=285\begin{aligned}& |\vec{a} \times \vec{c}|^2=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\1 & 2 & 3 \\3 & -2 & +4\end{array}\right|^2 \\\\& =|14 \hat{i}+5 \hat{j}-8 \hat{k}|^2 \\\\& =14^2+5^2+8^2=285\end{aligned}

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