Mathematics in Physics

JEE Main level, single correct, medium

50 of 121

Question 50 of 121

Let ∣A1⃗∣=3,∣A2⃗∣=5\left|\vec{{A}_{1}}\right|=3,\left|\vec{{A}_{2}}\right|=5 and ∣A1⃗+A2⃗∣=5.\left|\vec{{A}_{1}}+\vec{{A}_{2}}\right|=5. The value of (2A1⃗+3A2⃗)∙(3A1⃗−2A2⃗)\left(2\vec{{A}_{1}}+3\vec{{A}_{2}}\right)∙\left(3\vec{{A}_{1}}-2\vec{{A}_{2}}\right) is:

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Answer: (B) −118.5-118.5

∣A⃗1+A⃗2∣=5\left|{\vec{A}}_{1}+{\vec{A}}_{2}\right|=5
∣A⃗1+A⃗2∣2=25{\left|{\vec{A}}_{1}+{\vec{A}}_{2}\right|}^{2}=25
∣A⃗1∣2+∣A⃗2∣2+2A⃗1A⃗2=25{\left|{\vec{A}}_{1}\right|}^{2}+{\left|{\vec{A}}_{2}\right|}^{2}+2{\vec{A}}_{1}{\vec{A}}_{2}=25
9+25+2A⃗1∙A⃗2=259+25+2{\vec{A}}_{1}∙{\vec{A}}_{2}=25
A⃗1.A⃗2=−92{\vec{A}}_{1}.{\vec{A}}_{2}=-\dfrac{9}{2}
Now, (2A⃗1+3A⃗2)∙(3A⃗1−2A⃗2)=6∣A⃗1∣2−4A⃗1A⃗2+9A⃗1A⃗2−6∣A⃗2∣2\left(2{\vec{A}}_{1}+3{\vec{A}}_{2}\right)∙\left(3{\vec{A}}_{1}-2{\vec{A}}_{2}\right)=6{\left|{\vec{A}}_{1}\right|}^{2}-4{\vec{A}}_{1}{\vec{A}}_{2}+9{\vec{A}}_{1}{\vec{A}}_{2}-6{\left|{\vec{A}}_{2}\right|}^{2}
=6.∣A⃗1∣2+5A⃗1A⃗2−6∣A⃗2∣2=6.{\left|{\vec{A}}_{1}\right|}^{2}+5{\vec{A}}_{1}{\vec{A}}_{2}-6{\left|{\vec{A}}_{2}\right|}^{2}
=6×9+5(−92)−6×25=6\times 9+5\left(-\dfrac{9}{2}\right)-6\times 25
=54−452−150=54-\dfrac{45}{2}-150
=−118.5=-118.5

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