If overrightarrow{a} = 2widehat{i} - widehat{ | vedclass

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Question

(A) Given $\overrightarrow{a} = 2\widehat{i} - \widehat{j} + \widehat{k}$,$\overrightarrow{b} = \widehat{i} + \widehat{j} - 2\widehat{k}$,and $\overri

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(A) Given $\overrightarrow{a} = 2\widehat{i} - \widehat{j} + \widehat{k}$,$\overrightarrow{b} = \widehat{i} + \widehat{j} - 2\widehat{k}$,and $\overrightarrow{c} = \widehat{i} + 3\widehat{j} - (\lambda^2 + 3\lambda)\widehat{k}$.We are given that $\overrightarrow{a} \perp (\overrightarrow{c} - \lambda\overrightarrow{b})$,which implies $\overrightarrow{a} \cdot (\overrightarrow{c} - \lambda\overrightarrow{b}) = 0$.First,calculate $\overrightarrow{c} - \lambda\overrightarrow{b}$:$\overrightarrow{c} - \lambda\overrightarrow{b} = (\widehat{i} + 3\widehat{j} - (\lambda^2 + 3\lambda)\widehat{k}) - \lambda(\widehat{i} + \widehat{j} - 2\widehat{k})$$= (1 - \lambda)\widehat{i} + (3 - \lambda)\widehat{j} + (2\lambda - \lambda^2 - 3\lambda)\widehat{k}$$= (1 - \lambda)\widehat{i} + (3 - \lambda)\widehat{j} - (\lambda^2 + \lambda)\widehat{k}$.Now,compute the dot product $\overrightarrow{a} \cdot (\overrightarrow{c} - \lambda\overrightarrow{b}) = 0$:$(2\widehat{i} - \widehat{j} + \widehat{k}) \cdot ((1 - \lambda)\widehat{i} + (3 - \lambda)\widehat{j} - (\lambda^2 + \lambda)\widehat{k}) = 0$$2(1 - \lambda) - 1(3 - \lambda) + 1(-(\lambda^2 + \lambda)) = 0$$2 - 2\lambda - 3 + \lambda - \lambda^2 - \lambda = 0$$-\lambda^2 - 2\lambda - 1 = 0$$\lambda^2 + 2\lambda + 1 = 0$$(\lambda + 1)^2 = 0$$\lambda = -1$.Since there is only one value for $\lambda$,the sum of different values is $-1$.

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