If overline{a}=2 hat{i}+2 hat{j}+3 hat{k},ove | vedclass

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(D) Given that $\overline{b}+\lambda \overline{a}$ is perpendicular to $\overline{c}$,their dot product must be zero: $(\overline{b}+\lambda \overline

Vedclass · Accepted Answer

$\frac{1}{8}$

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

If $\overline{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}$,$\overline{b}=-\hat{i}+2 \hat{j}+\hat{k}$ and $\overline{c}=3 \hat{i}+\hat{j}$ such that $\overline{b}+\lambda \overline{a}$ is perpendicular to $\overline{c}$,then $\lambda$ is

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