Let overrightarrow{a}=2 hat{i}-hat{j}+3 hat{k | vedclass

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(C) Given $\overrightarrow{a}=2 \hat{i}-\hat{j}+3 \hat{k}$ and $\overrightarrow{b}=3 \hat{i}-5 \hat{j}+\hat{k}$.From $\overrightarrow{a} 	imes \overr

Vedclass · Accepted Answer

$308$

Vedclass · Answer

(C) Given $\overrightarrow{a}=2 \hat{i}-\hat{j}+3 \hat{k}$ and $\overrightarrow{b}=3 \hat{i}-5 \hat{j}+\hat{k}$.From $\overrightarrow{a} 	imes \overrightarrow{c}=\overrightarrow{c} 	imes \overrightarrow{b}$,we have $\overrightarrow{a} 	imes \overrightarrow{c}+\overrightarrow{b} 	imes \overrightarrow{c}=0$,which implies $(\overrightarrow{a}+\overrightarrow{b}) 	imes \overrightarrow{c}=0$.This means $\overrightarrow{c}$ is parallel to $(\overrightarrow{a}+\overrightarrow{b})$.Let $\overrightarrow{c}=\lambda(\overrightarrow{a}+\overrightarrow{b})=\lambda(5 \hat{i}-6 \hat{j}+4 \hat{k})$.Then $|\overrightarrow{c}|^2=\lambda^2(5^2+(-6)^2+4^2)=77 \lambda^2$.Given $(\overrightarrow{a}+\overrightarrow{c}) \cdot(\overrightarrow{b}+\overrightarrow{c})=168$.Expanding this,we get $\overrightarrow{a} \cdot \overrightarrow{b}+\overrightarrow{a} \cdot \overrightarrow{c}+\overrightarrow{c} \cdot \overrightarrow{b}+|\overrightarrow{c}|^2=168$.Calculating $\overrightarrow{a} \cdot \overrightarrow{b} = (2)(3)+(-1)(-5)+(3)(1) = 6+5+3=14$.Substituting $\overrightarrow{c}=\lambda(\overrightarrow{a}+\overrightarrow{b})$,we get $14+\lambda(\overrightarrow{a}+\overrightarrow{b}) \cdot (\overrightarrow{a}+\overrightarrow{b})+77 \lambda^2=168$.$14+\lambda|\overrightarrow{a}+\overrightarrow{b}|^2+77 \lambda^2=168$.$14+\lambda(77)+77 \lambda^2=168$.$77 \lambda^2+77 \lambda-154=0 \Rightarrow \lambda^2+\lambda-2=0$.Solving for $\lambda$,we get $(\lambda+2)(\lambda-1)=0$,so $\lambda=-2$ or $\lambda=1$.Since $|\overrightarrow{c}|^2=77 \lambda^2$,the maximum value occurs at $\lambda=-2$.$|\overrightarrow{c}|^2=77 	imes (-2)^2 = 77 	imes 4 = 308$.

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