If bar{a} and bar{b} are vectors such that |b | vedclass

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(A) Given $\vec{a} 	imes(2 \hat{i}+3 \hat{j}+4 \hat{k})=(2 \hat{i}+3 \hat{j}+4 \hat{k}) 	imes \vec{b}$.We know that $\vec{u} 	imes \vec{v} = -\vec{

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$4$

Vedclass · Answer

(A) Given $\vec{a} 	imes(2 \hat{i}+3 \hat{j}+4 \hat{k})=(2 \hat{i}+3 \hat{j}+4 \hat{k}) 	imes \vec{b}$.We know that $\vec{u} 	imes \vec{v} = -\vec{v} 	imes \vec{u}$,so $(2 \hat{i}+3 \hat{j}+4 \hat{k}) 	imes \vec{b} = -\vec{b} 	imes (2 \hat{i}+3 \hat{j}+4 \hat{k})$.Substituting this into the equation,we get $\vec{a} 	imes(2 \hat{i}+3 \hat{j}+4 \hat{k}) + \vec{b} 	imes(2 \hat{i}+3 \hat{j}+4 \hat{k}) = \vec{0}$.This implies $(\vec{a}+\vec{b}) 	imes(2 \hat{i}+3 \hat{j}+4 \hat{k}) = \vec{0}$.This means the vector $(\vec{a}+\vec{b})$ is parallel to $(2 \hat{i}+3 \hat{j}+4 \hat{k})$.Let $\vec{a}+\vec{b} = \lambda(2 \hat{i}+3 \hat{j}+4 \hat{k})$ for some scalar $\lambda$.Taking the magnitude on both sides,$|\vec{a}+\vec{b}| = |\lambda| \sqrt{2^2+3^2+4^2} = |\lambda| \sqrt{4+9+16} = |\lambda| \sqrt{29}$.Given $|\vec{a}+\vec{b}| = \sqrt{29}$,we have $|\lambda| \sqrt{29} = \sqrt{29}$,which gives $\lambda = \pm 1$.Thus,$\vec{a}+\vec{b} = \pm(2 \hat{i}+3 \hat{j}+4 \hat{k})$.Now,calculate the dot product: $(\vec{a}+\vec{b}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k}) = \pm(2 \hat{i}+3 \hat{j}+4 \hat{k}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k})$.$= \pm((2)(-7) + (3)(2) + (4)(3)) = \pm(-14 + 6 + 12) = \pm 4$.Therefore,a possible value is $4$.

If $\bar{a}$ and $\bar{b}$ are vectors such that $|\bar{a}+\bar{b}|=\sqrt{29}$ and $\bar{a} \times(2 \hat{i}+3 \hat{j}+4 \hat{k})=(2 \hat{i}+3 \hat{j}+4 \hat{k}) \times \bar{b}$,then a possible value of $(\bar{a}+\bar{b}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k})$ is

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