Between the following two statements :Stateme | vedclass

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(B) For Statement $-I$:Given $\vec{a} = \hat{i}+2\hat{j}-3\hat{k}$ and $\vec{b} = 2\hat{i}+\hat{j}-\hat{k}$.$\vec{a} 	imes \vec{r} = \vec{a} 	imes \

Vedclass · Accepted Answer

Statement $-I$ is incorrect but Statement $-II$ is correct.

Vedclass · Answer

(B) For Statement $-I$:Given $\vec{a} = \hat{i}+2\hat{j}-3\hat{k}$ and $\vec{b} = 2\hat{i}+\hat{j}-\hat{k}$.$\vec{a} 	imes \vec{r} = \vec{a} 	imes \vec{b} \implies \vec{a} 	imes (\vec{r}-\vec{b}) = \vec{0}$.This implies $\vec{r}-\vec{b} = k\vec{a}$ for some scalar $k$.So,$\vec{r} = \vec{b} + k\vec{a}$.Given $\vec{a} \cdot \vec{r} = 0$,so $\vec{a} \cdot (\vec{b} + k\vec{a}) = 0$.$\vec{a} \cdot \vec{b} + k|\vec{a}|^2 = 0$.$\vec{a} \cdot \vec{b} = (1)(2) + (2)(1) + (-3)(-1) = 2+2+3 = 7$.$|\vec{a}|^2 = 1^2+2^2+(-3)^2 = 1+4+9 = 14$.$7 + 14k = 0 \implies k = -\frac{1}{2}$.$\vec{r} = \vec{b} - \frac{1}{2}\vec{a} = (2\hat{i}+\hat{j}-\hat{k}) - \frac{1}{2}(\hat{i}+2\hat{j}-3\hat{k}) = \frac{3}{2}\hat{i} + 0\hat{j} + \frac{1}{2}\hat{k}$.Magnitude $|\vec{r}| = \sqrt{(\frac{3}{2})^2 + 0^2 + (\frac{1}{2})^2} = \sqrt{\frac{9}{4} + \frac{1}{4}} = \sqrt{\frac{10}{4}} = \frac{\sqrt{10}}{2}$.Since $\frac{\sqrt{10}}{2} 
eq \sqrt{10}$,Statement $-I$ is incorrect.For Statement $-II$:In $	riangle ABC$,$\cos 2A + \cos 2B + \cos 2C = -1 - 4\cos A \cos B \cos C$.For an acute triangle,$\cos A, \cos B, \cos C > 0$,so $\cos 2A + \cos 2B + \cos 2C < -1$. However,the inequality $\cos 2A + \cos 2B + \cos 2C \geq -\frac{3}{2}$ is a standard result for any triangle. Thus,Statement $-II$ is correct.

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