If vec{a} = 2hat{i} - hat{j} + hat{k},vec{b} | vedclass

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(A) Given $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$,$\vec{b} = \hat{i} + \hat{j} - 2\hat{k}$,and $\vec{c} = \hat{i} + 3\hat{j} - (\lambda^2 + 3\lambda)

Vedclass · Accepted Answer

$-1$

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(A) Given $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$,$\vec{b} = \hat{i} + \hat{j} - 2\hat{k}$,and $\vec{c} = \hat{i} + 3\hat{j} - (\lambda^2 + 3\lambda)\hat{k}$.We are given that $\vec{a}$ is perpendicular to $\vec{c} - \lambda\vec{b}$,so $\vec{a} \cdot (\vec{c} - \lambda\vec{b}) = 0$.First,calculate $\vec{c} - \lambda\vec{b}$:$\vec{c} - \lambda\vec{b} = (\hat{i} + 3\hat{j} - (\lambda^2 + 3\lambda)\hat{k}) - \lambda(\hat{i} + \hat{j} - 2\hat{k})$$= (1 - \lambda)\hat{i} + (3 - \lambda)\hat{j} + (2\lambda - \lambda^2 - 3\lambda)\hat{k}$$= (1 - \lambda)\hat{i} + (3 - \lambda)\hat{j} - (\lambda^2 + \lambda)\hat{k}$.Now,take the dot product with $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$:$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 $-1$.

If $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$,$\vec{b} = \hat{i} + \hat{j} - 2\hat{k}$,and $\vec{c} = \hat{i} + 3\hat{j} - (\lambda^2 + 3\lambda)\hat{k}$ (where $\lambda$ is a constant) and $\vec{a}$ is perpendicular to $\vec{c} - \lambda\vec{b}$,then the sum of different values of $\lambda$ is:

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