If the vectors vec{a}=2 hat{i}+2 hat{j}+3 hat | vedclass

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(C) Given that $(\vec{a}+\lambda \vec{b})$ is perpendicular to $\vec{c}$,their dot product must be zero: $(\vec{a}+\lambda \vec{b}) \cdot \vec{c} = 0$

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(C) Given that $(\vec{a}+\lambda \vec{b})$ is perpendicular to $\vec{c}$,their dot product must be zero: $(\vec{a}+\lambda \vec{b}) \cdot \vec{c} = 0$ First,calculate $(\vec{a}+\lambda \vec{b})$: $\vec{a}+\lambda \vec{b} = (2 \hat{i}+2 \hat{j}+3 \hat{k}) + \lambda(-\hat{i}+2 \hat{j}+\hat{k}) = (2-\lambda) \hat{i} + (2+2\lambda) \hat{j} + (3+\lambda) \hat{k}$ Now,take the dot product with $\vec{c} = 3 \hat{i} + \hat{j} + 0 \hat{k}$: $((2-\lambda) \hat{i} + (2+2\lambda) \hat{j} + (3+\lambda) \hat{k}) \cdot (3 \hat{i} + \hat{j} + 0 \hat{k}) = 0$ $3(2-\lambda) + 1(2+2\lambda) + 0(3+\lambda) = 0$ $6 - 3\lambda + 2 + 2\lambda = 0$ $8 - \lambda = 0$ $\lambda = 8$

If the vectors $\vec{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}$,$\vec{b}=-\hat{i}+2 \hat{j}+\hat{k}$ and $\vec{c}=3 \hat{i}+\hat{j}$ are such that $(\vec{a}+\lambda \vec{b})$ is perpendicular to $\vec{c}$,then the value of $\lambda$ is

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