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QJEEM24Maa5cf: JEE Main 2024 Physics - Class 11 JEEM24Maa5cf Solution: Let and . If is a is vector such that and the angle between and …

Let a→=6i^+j^−k^\overrightarrow{\mathrm{a}}=6 \hat{i}+\hat{j}-\hat{k} and b→=i^+j^\overrightarrow{\mathrm{b}}=\hat{i}+\hat{j} . If c→\overrightarrow{\mathrm{c}} is a is vector such that ∣c→∣≥6,a→⋅c→=6∣c→∣,∣c→−a→∣=22|\overrightarrow{\mathrm{c}}| \geq 6, \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=6|\overrightarrow{\mathrm{c}}|,|\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}}|=2 \sqrt{2} and the angle between a⃗×b⃗\vec{a} \times \vec{b} and c⃗\vec{c} is 60∘60^{\circ} , then ∣(a⃗×b⃗)×c⃗∣|(\vec{a} \times \vec{b}) \times \vec{c}| is equal to:

Multiple Choice Options

  • A 326\frac{3}{2} \sqrt{6}
  • B 92(6−6)\frac{9}{2}(6-\sqrt{6})
  • C 92(6+6)\frac{9}{2}(6+\sqrt{6})
  • D 323\frac{3}{2} \sqrt{3}

Step-by-Step Verified Solution

amp;∣(a⃗×b⃗)×c⃗∣=∣a⃗×b⃗∣∣c⃗∣sin⁡60∘amp;∣iamp;jamp;k6amp;1amp;−11amp;1amp;0∣=i(1)−j(1)+k(5)amp;=i−j+5kamp;∣a⃗×b⃗∣=1+1+25=27amp;∣c⃗−a⃗∣=22amp;c2+a2−2ac=8amp;c2−12c+30=0amp;c=12+242=6+6amp;⇒∣(a⃗×b⃗)×c⃗∣=27×(6+6)×32amp;=a2(6+6)\begin{aligned}& |(\vec{a} \times \vec{b}) \times \vec{c}|=|\vec{a} \times \vec{b}||\vec{c}| \sin 60^{\circ} \\& \left|\begin{array}{ccc}i & j & k \\6 & 1 & -1 \\1 & 1 & 0\end{array}\right|=i(1)-j(1)+k(5) \\& =i-j+5 k \\& |\vec{a} \times \vec{b}|=\sqrt{1+1+25}=\sqrt{27} \\& |\vec{c}-\vec{a}|=2 \sqrt{2} \\& c^2+a^2-2 a c=8 \\& c^2-12 c+30=0 \\& c=\frac{12+\sqrt{24}}{2}=6+\sqrt{6} \\& \Rightarrow|(\vec{a} \times \vec{b}) \times \vec{c}|=\sqrt{27} \times(6+\sqrt{6}) \times \frac{\sqrt{3}}{2} \\& =\frac{a}{2}(6+\sqrt{6}) \\\end{aligned}

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