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QJEEM23Ma7041: JEE Main 2023 Physics - Class 11 JEEM23Ma7041 Solution: Let be three vectors such that and . If the angle between and…

Let a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} be three vectors such that

∣a⃗∣=31,4∣b⃗∣=∣c⃗∣=2|\vec{a}|=\sqrt{31}, 4|\vec{b}|=|\vec{c}|=2 and 2(a⃗×b⃗)=3(c⃗×a⃗)2(\vec{a} \times \vec{b})=3(\vec{c} \times \vec{a}) .

If the angle between b⃗\vec{b} and c⃗\vec{c} is 2π3\frac{2 \pi}{3} , then (a⃗×c⃗a⃗⋅b⃗)2\left(\frac{\vec{a} \times \vec{c}}{\vec{a} \cdot \vec{b}}\right)^{2} is equal to __________.

Step-by-Step Verified Solution

2(a⃗×b⃗)=3(c⃗×a⃗)2(\vec{a} \times \vec{b})=3(\vec{c} \times \vec{a})

a⃗×(2b⃗+3c⃗)=0\vec{a} \times(2 \vec{b}+3 \vec{c})=0

amp;a⃗=λ(2b⃗+3c⃗)amp;∣a⃗∣2=λ2(4∣b∣2+9∣c∣2+12b⃗⋅c⃗)amp;31=31λ2amp;λ=±1amp;a⃗=±(2b⃗+3c⃗)amp;∣a⃗×c⃗∣∣a⃗⋅b⃗∣=2∣b⃗×c⃗∣2b⃗⋅b⃗+3c⃗⋅b⃗amp;∣b⃗×c⃗∣2=14⋅4−(1−12)2amp;=34amp;∴∣a⃗×c⃗∣∣a⃗⋅b⃗∣=32⋅14−32=3−1amp;(∣a⃗×c⃗∣∣a⃗⋅b⃗∣)2=3\begin{aligned}& \vec{a}=\lambda(2 \vec{b}+3 \vec{c}) \\\\& |\vec{a}|^{2}=\lambda^{2}\left(4|b|^{2}+9|c|^{2}+12 \vec{b} \cdot \vec{c}\right) \\\\& 31=31 \lambda^{2} \\\\& \lambda=\pm 1 \\\\& \vec{a}=\pm(2 \vec{b}+3 \vec{c}) \\\\& \frac{|\vec{a} \times \vec{c}|}{|\vec{a} \cdot \vec{b}|}=\frac{2|\vec{b} \times \vec{c}|}{2 \vec{b} \cdot \vec{b}+3 \vec{c} \cdot \vec{b}} \\\\& |\vec{b} \times \vec{c}|^{2}=\frac{1}{4} \cdot 4-\left(1-\frac{1}{2}\right)^{2} \\\\& =\frac{3}{4} \\\\& \therefore \frac{|\vec{a} \times \vec{c}|}{|\vec{a} \cdot \vec{b}|}=\frac{\sqrt{3}}{2 \cdot \frac{1}{4}-\frac{3}{2}}=\frac{\sqrt{3}}{-1} \\\\& \left(\frac{|\vec{a} \times \vec{c}|}{|\vec{a} \cdot \vec{b}|}\right)^{2}=3\end{aligned}

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