If bar{a}=2 hat{i}+3 hat{j}+2 hat{k},bar{b}=2 | vedclass

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(D) According to the given condition,the vector $(\bar{a}+\lambda \bar{b})$ is perpendicular to $\bar{c}$,so their dot product must be zero: $(\bar{a}

Vedclass · Accepted Answer

$\frac{-3}{5}$

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(D) According to the given condition,the vector $(\bar{a}+\lambda \bar{b})$ is perpendicular to $\bar{c}$,so their dot product must be zero: $(\bar{a}+\lambda \bar{b}) \cdot \bar{c} = 0$ First,calculate $\bar{a}+\lambda \bar{b}$: $\bar{a}+\lambda \bar{b} = (2 \hat{i}+3 \hat{j}+2 \hat{k}) + \lambda(2 \hat{i}+\hat{j}-\hat{k}) = (2+2\lambda) \hat{i} + (3+\lambda) \hat{j} + (2-\lambda) \hat{k}$ Now,take the dot product with $\bar{c} = 3 \hat{i} - \hat{j} + 0 \hat{k}$: $[(2+2\lambda) \hat{i} + (3+\lambda) \hat{j} + (2-\lambda) \hat{k}] \cdot (3 \hat{i} - \hat{j} + 0 \hat{k}) = 0$ $3(2+2\lambda) - 1(3+\lambda) + 0(2-\lambda) = 0$ $6 + 6\lambda - 3 - \lambda = 0$ $3 + 5\lambda = 0$ $5\lambda = -3$ $\lambda = \frac{-3}{5}$

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

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