Let vec{a}, vec{b} and vec{c} be three unit v | vedclass

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(A) Given $\vec{a}+\vec{b}+\vec{c}=\vec{0}$ and $|\vec{a}|=|\vec{b}|=|\vec{c}|=1$.Squaring both sides: $(\vec{a}+\vec{b}+\vec{c}) \cdot (\vec{a}+\vec{

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$\left(-\frac{3}{2}, 3 \vec{a} 	imes \vec{b}ight)$

Vedclass · Answer

(A) Given $\vec{a}+\vec{b}+\vec{c}=\vec{0}$ and $|\vec{a}|=|\vec{b}|=|\vec{c}|=1$.Squaring both sides: $(\vec{a}+\vec{b}+\vec{c}) \cdot (\vec{a}+\vec{b}+\vec{c}) = 0$.$|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}) = 0$.$1+1+1+2\lambda = 0 \Rightarrow 3+2\lambda = 0 \Rightarrow \lambda = -\frac{3}{2}$.Now,$\vec{a}+\vec{b}+\vec{c}=\vec{0} \Rightarrow \vec{a}+\vec{b}=-\vec{c}$.Taking cross product with $\vec{a}$: $\vec{a} 	imes (\vec{a}+\vec{b}) = \vec{a} 	imes (-\vec{c}) \Rightarrow \vec{a} 	imes \vec{b} = \vec{c} 	imes \vec{a}$.Similarly,taking cross product with $\vec{b}$: $(\vec{a}+\vec{b}) 	imes \vec{b} = (-\vec{c}) 	imes \vec{b} \Rightarrow \vec{a} 	imes \vec{b} = \vec{b} 	imes \vec{c}$.Thus,$\vec{a} 	imes \vec{b} = \vec{b} 	imes \vec{c} = \vec{c} 	imes \vec{a}$.Therefore,$\vec{d} = \vec{a} 	imes \vec{b} + \vec{a} 	imes \vec{b} + \vec{a} 	imes \vec{b} = 3(\vec{a} 	imes \vec{b})$.The ordered pair is $\left(-\frac{3}{2}, 3 \vec{a} 	imes \vec{b}ight)$.

Column $I$	Column $II$
$(A)$ If $\vec{a}=\hat{j}+\sqrt{3} \hat{k}, \vec{b}=-\hat{j}+\sqrt{3} \hat{k}$ and $\vec{c}=2 \sqrt{3} \hat{k}$ form a triangle,then the internal angle of the triangle between $\vec{a}$ and $\vec{b}$ is	$(p)$ $\frac{\pi}{6}$
$(B)$ If $\int_a^b(f(x)-3 x) d x=a^2-b^2$,then the value of $f\left(\frac{\pi}{6}\right)$ is	$(q)$ $\frac{2 \pi}{3}$
$(C)$ The value of $\frac{\pi^2}{\ln 3} \int_{1 / 6}^{5 / 6} \sec (\pi x) d x$ is	$(r)$ $\frac{\pi}{3}$
$(D)$ The maximum value of $\|\operatorname{Arg}(\frac{1}{1-z})\|$ for $\|z\|=1, z \neq 1$ is given by	$(s)$ $\pi$
	$(t)$ $\frac{\pi}{2}$

List-$I$	List-$II$
$(i)$ $\overrightarrow{a} \cdot \overrightarrow{b}$	$(A)$ $\overrightarrow{a} \cdot \overrightarrow{d}$
(ii) $\overrightarrow{b} \cdot \overrightarrow{c}$	$(B)$ $3$
(iii) $[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]$	$(C)$ $\overrightarrow{b} \cdot \overrightarrow{d}$
(iv) $\overrightarrow{b} \times \overrightarrow{c}$	$(D)$ $2\hat{i}-2\hat{k}$
	$(E)$ $2\hat{j}+2\hat{k}$
	$(F)$ $4$

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